[置顶] 【主席树】 跟岛娘学主席树
A http://www.spoj.com/problems/DQUERY/
岛娘的资料：
http://paste.ubuntu.com/1426159/
http://paste.ubuntu.com/1426225/
http://paste.ubuntu.com/1426225/
题意：
给一个序列，求某段区间内的不同元素个数。
解法：
1、离线 + 树状数组
将询问按照左端排序，维护元素第一次出现位置，树状数组。
const int N = 40000;
int tree[N] , a[N];
int n;
void add(int x){
for (; x < N ; x += low_bit(x)){
tree[x] ++;
}
}
int query(int x){
int ret = 0;
for( ; x ; x -= low_bit(x)){
ret += tree[x];
}
return ret;
}
struct IO{
int l , r , id;
void input(){
RD(l , r);
}
bool operator < (const IO & A) const{
return l < A.l;
}
}Q[200001];
int ans[200001];
typedef map<int , int> MI;
MI P;
MI :: iterator iter;
set<int> fir;
int nex[40000] , q;
void solve(){
for (int i = 1 ; i <= n ; ++i)
RD(a[i]);
P.clear();
for(int i = n ; i ; --i){
iter = P.find(a[i]);
if (iter == P.end()){
nex[i] = n + 1;
}
else nex[i] = iter -> se;
P[a[i]] = i;
};
RD(q);
for (int i = 0 ; i < q ; ++i){
Q[i].input();
Q[i].id = i;
}
sort(Q , Q + q);
RST(tree);
fir.clear();
for (int i = 1 ; i <= n ; ++i){
if (fir.find(a[i]) == fir.end()){
fir.insert(a[i]);
add(i);
}
}
int head = 1 ;
for (int i = 0 ; i < q ; ++i){
while(head < Q[i].l){
// cout << nex[head] << endl;
if(nex[head] <= n) add(nex[head]);
head++;
}
ans[Q[i].id] = query(Q[i].r);
ans[Q[i].id] -= query(Q[i].l - 1);
}
for (int i = 0 ; i < q ; ++i)
OT(ans[i]);
}
int main(){
while(~scanf("%d" , &n)) solve();
}
2、在线 + 主席树
不用离线，对于每个起始位置维护一颗主席树。每次询问在线即可
TYPE 1 ：
const int N = 30009;
int a[N];
MII hash;
MII IT iter;
int n;
namespace Fotile_Tree{
const int FN = 1000009;
int l[FN] /*lson*/, r[FN] , c[FN]/*Value*/ , tot;
int T[FN];//T 主席树head
int build(int L , int R){
int rt = ++tot ;
if (L >= R) return rt;
int mid = L + R >> 1;
l[rt] = build(L , mid);
r[rt] = build(mid + 1 , R);
return rt;
}
int update(int rt , int p , int d){
int newrt = ++tot , root = newrt;
c[newrt] = c[rt] + d ;
int L = 1 , R = n;
int mid;
while(L < R){
mid = L + R >> 1;
if (p <= mid){
l[newrt] = ++tot , r[newrt] = r[rt];
newrt = l[newrt] , rt = l[rt] , R = mid;
}
else{
l[newrt] = l[rt] , r[newrt] = ++tot;
newrt = r[newrt] , rt = r[rt] , L = mid + 1;
}
c[newrt] = c[rt] + d;
}
return root;
}
int query(int rt , int p){ //Query prefix
int res = 0 , L = 1 , R = n;
int mid ;
while(p != R){
mid = L + R >> 1;
if (p <= mid){
rt = l[rt];
R = mid;
}
else{
res += c[l[rt]];
rt = r[rt];
L = mid + 1;
}
}
return res + c[rt];
}
} using namespace Fotile_Tree;
int nex;
void solve(){
for (int i = 1 ; i <= n; ++i) RD(a[i]);
hash.clear();
T[n + 1] = build(1 , n);
DWN_1(i , n , 1){
iter = hash.find(a[i]);
if (iter == hash.end()) nex = 0;
else nex = iter -> se;
if (nex){
int u = update(T[i + 1] , nex , -1);
T[i] = update(u , i , 1);
}
else
T[i] = update(T[i + 1] , i , 1);
hash[a[i]] = i;
}
Rush{
int ql , qr;
RD(ql , qr);
OT(query(T[ql] , qr));
}
}
int main(){
while(~scanf("%d" , &n)) solve();
}
TYPE 2：
const int N = 30009;
int a[N];
MII hash;
MII IT iter;
int n;
namespace Fotile_Tree{
const int FN = 1000009;
int l[FN] , r[FN] , c[FN];
int T[N] , tot , _p , _c = 0;
int new_node(){ return ++tot ;}
void pushup(int rt) {
c[rt] = c[l[rt]] + c[r[rt]];
}
int node(){
int rt = new_node();
c[rt] = _c;
return rt;
}
int node(int a , int b){
int rt = new_node();
l[rt] = a , r[rt] = b;
pushup(rt);
return rt;
}
int path(int rt , int L = 1 , int R = n){
if (L == R) return node();
int mid = L + R >> 1;
if (_p <= mid)
return node(path(l[rt] , L , mid) , r[rt]);
else
return node(l[rt] , path(r[rt] , mid + 1 , R));
}
int new_path(int rt , int p , int c){
_p = p , _c = c;
return path(rt);
}
void build(int &rt ,int L = 1 , int R = n){
rt = new_node();
if (L == R) return;
int mid = L + R >> 1;
build(l[rt] , L , mid);
build(r[rt] , mid + 1 , R);
pushup(rt);
}
inline int query(int rt , int p){
int res = 0 , L = 1 , R = n;
while(p != R){
int mid = L + R >> 1;
if (p <= mid){
rt = l[rt] , R = mid;
}
else{
res += c[l[rt]];
rt = r[rt] , L = mid + 1;
}
}
return res + c[rt];
}
} using namespace Fotile_Tree;
int nex;
void solve(){
for (int i = 1 ; i <= n; ++i) RD(a[i]);
hash.clear();
build(T[n + 1] , 1 , n);
DWN_1(i , n , 1){
iter = hash.find(a[i]);
if (iter == hash.end()) nex = 0;
else nex = iter -> se;
if (nex){
int u = new_path(T[i + 1] , nex , 0);
T[i] = new_path(u , i , 1);
}
else
T[i] = new_path(T[i + 1] , i , 1);
hash[a[i]] = i;
}
Rush{
int ql , qr;
RD(ql , qr);
OT(query(T[ql] , qr));
}
}
int main(){
while(~scanf("%d" , &n)) solve();
}
B POJ 2104 && 2761
题意：
静态求区间第K 大数
方法：
1、树的第i位记录第i大的树在 suffix 后面出现了多少次。
2、从后往前（DWN n -> 1 建树） 每次单点更新
3、每次询问的时候，T[l] - T[r + 1] 即可，如果 个数 >= k 递归左边，否则递归右边
int n , Q;
const int N = 1e5 + 9;
int a[N];
namespace HASH{
int h[N] , m;
void hashinit(){
for (int i = 1 ; i <= n ; ++i)
h[i] = a[i];
h[0] = -INF;
sort(h + 1 , h + n + 1);
m = unique(h , h + n + 1) - h - 1;
}
int hashQ(int x){
return lower_bound(h , h + m + 1 , x) - h;
}
int number(int x){
// cout << x << endl;
return h[x];
}
}using namespace HASH;
namespace Persistent_tree{
const int PN = 10000009;
int T[N];
int l[PN] , r[PN] , c[PN] , tot;
int build(int L , int R){
int rt = ++tot;
if (L >= R) return rt;
int mid = L + R >> 1;
l[rt] = build(L , mid);
r[rt] = build(mid + 1 , R);
c[rt] = 0;
return rt;
}
int update(int rt , int p , int d){
int nrt = ++tot , root = nrt;
c[nrt] = c[rt] + d;
int L = 1 , R = m;
int mid ;
while (L < R){
mid = L + R >> 1;
if (p <= mid){
l[nrt] = ++tot , r[nrt] = r[rt];
nrt = l[nrt] , rt = l[rt] , R = mid;
}
else{
l[nrt] = l[rt] , r[nrt] = ++ tot;
nrt = r[nrt] , rt = r[rt] , L = mid +1;
}
c[nrt] = c[rt] + d;
}
return root;
}
int query(int rtl , int rtr , int k){
int L = 1 , R = m;
int mid;
while(L < R){
mid = L + R >> 1;
if (c[l[rtl]] - c[l[rtr]] >= k){
rtl = l[rtl];
rtr = l[rtr];
R = mid;
}
else{
k -= c[l[rtl]] - c[l[rtr]];
rtl = r[rtl];
rtr = r[rtr];
L = mid + 1;
}
// printf("[%d,%d] %d\n" , h[L] , h[R] , k);
}
assert(L == R);
return L;
}
}using namespace Persistent_tree;
void solve(){
for(int i = 1 ; i <= n ; ++i)
RD(a[i]);
hashinit();
tot = 0;
T[n + 1] = build(1 , m);
DWN_1(i , n , 1){
int now = hashQ(a[i]);
// printf("%d = %d\n" ,a[i] , now);
T[i] = update(T[i + 1] , now , 1);
}
DO(Q){
int L , R , k;
RD(L , R , k);
OT(number(query(T[L] , T[R + 1] , k)));
}
}
int main(){
while(~scanf("%d%d" , &n , &Q)) solve();
}
C http://www.spoj.com/problems/COT/
岛岛的资料：
题意：
给一颗树（节点 <= 1e5），树上每个点都有一个权值。
给Q个询问（Q<=1e5），每次给出x , y , k 求 x y 的链上面第k大是多少。
方法：
跟kth Number 差不多，只不过这次是按照Tree建立的Persistent_tree。之前要预处理一下LCA。
所以整个题目的解法就是Hash + LCA + Persistent_tree。
细节：
1、需要Tree[x] + Tree[y] - 2 * Tree[lca] + lca那个点是否在区间内。
2、Hash 的时候范围弄错了T_T
给一组好数据哈：
Input
8 6 105 2 9 3 8 5 7 7 1 2 1 3 1 4 3 5 3 6 3 7 4 8 2 5 1 2 5 2 2 5 3 2 5 4 7 8 2 8 8 1 13 1 7 3 6 1 2 9 4 9 5 8 7 7 4 1 2 1 3 1 8 2 4 2 5 3 6 3 7 8 9 8 10 5 13 6 12 7 11 12 11 5Output：
2 8 9 105 7 7 9
Debug 了一夜之后大叹一口气终于过了。。
尼玛还卡代码长度！！！！太不科学了！！
我的Code 用三个namespace 罩起来，应该看起来清晰大气上档次！！
/**
纪念一下
![[置顶] 【主席树】 跟岛娘学主席树_第1张图片](http://img.e-com-net.com/image/info5/6cff6055478e4f1981ba92b82b82f7a1.png)
*/
#define LOCAL
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
using namespace std;
#define SZ(A) int(A.size())
#define PB push_back
typedef vector<int> VI;
#define REP(i, n) for (int i=0;i<int(n);++i)
#define REP_1(i, n) for (int i=1;i<=int(n);++i)
template<class T> inline int RD(T &x){
char c; for (c = getchar(); c < '0'; c = getchar()); x = c - '0'; for (c = getchar(); c >= '0'; c = getchar()) x = x * 10 + c - '0';
return x;
}
/* .................................................................................................................................. */
const int N = 1e5 + 19;
int a[N] , n;
namespace HASH{
int h_[N] , m;
void hashInit(){
for(int i = 1 ; i <= n ; ++i){
h_[i] = a[i];
}
sort(h_ + 1 , h_ + n + 1);
m = unique(h_ + 1 , h_ + n + 1) - h_ - 1;
}
int hashQ(int x){
return lower_bound(h_ + 1, h_ + m + 1 , x) - h_;
}
int oQ(int x){
return h_[x];
}
}using namespace HASH;
VI edge[N];
namespace LCA{
int pw[30];
int dep[N];
int pa[N][30] , pre[N];
void initLCA(){
for (int i = 0 ; i < 30 ; ++i)
pw[i] = 1 << i;
}
void dfs(int u , int v , int deep){
dep[u] = deep;
pre[u] = v;
for(int i = 0 ; i < SZ(edge[u]) ; ++i){
int go = edge[u][i];
if (go == v) continue;
dfs(go , u , deep + 1);
}
}
void make_parent(){
memset(pa , -1 , sizeof(pa));
REP_1(i , n)
pa[i][0] = pre[i];
for (int j = 1 ; pw[j] < n ; j++)
REP_1(i , n)
if (pa[i][j - 1] != -1)
pa[i][j] = pa[pa[i][j - 1]][j - 1];
}
int lca(int x , int y){
int i , log = 0;
if (dep[x] < dep[y]) swap(x , y);
while(pw[log + 1] <= dep[x]) log++;
for (i = log ; i >= 0 ; --i)
if (dep[x] - pw[i] >= dep[y])
x = pa[x][i];
if (x == y) return x;
for (i = log ; i >= 0 ; i--)
if (pa[x][i] != -1 && pa[x][i] != pa[y][i])
x = pa[x][i] , y = pa[y][i];
return pre[x];
}
}using namespace LCA;
namespace Persistent_tree{
const int TN = N * 23;
int T[N];
int l[TN] , r[TN] , c[TN] , tot;
int build(int L , int R){
int rt = ++ tot;
if (L >= R) return rt;
int mid = L + R >> 1;
l[rt] = build(L , mid);
r[rt] = build(mid + 1 , R);
c[rt] = 0;
return rt;
}
int update(int rt , int p , int d){
int nrt = ++tot , root = nrt;
c[nrt] = c[rt] + d;
int L = 1 , R = m;
int mid ;
while (L < R){
mid = L + R >> 1;
if (p <= mid){
l[nrt] = ++ tot , r[nrt] = r[rt];
nrt = l[nrt] , rt = l[rt] , R = mid;
}
else{
l[nrt] = l[rt] , r[nrt] = ++tot;
nrt = r[nrt] , rt = r[rt] , L = mid + 1;
}
c[nrt] = c[rt] + d;
}
return root;
}
int query(int x , int y , int z , int k){
int L = 1 , R = m;
int mid;
int rtx = T[x] , rty = T[y] , rtlca = T[z];
int now = a[z] , t;
while(L < R){
mid = L + R >> 1;
// printf("C %d %d %d\n" , c[l[rtx]] , c[l[rty]] , c[l[rtlca]]);
if ((t = c[l[rtx]] + c[l[rty]] - 2 * c[l[rtlca]] + (L <= now && now <= mid) )>= k){
rtx = l[rtx];
rty = l[rty];
rtlca = l[rtlca];
R = mid;
}
else{
k -= t;
rtx = r[rtx];
rty = r[rty];
rtlca = r[rtlca];
L = mid + 1;
}
// printf("%d %d - [%d %d] %d\n" , L , R , oQ(L) , oQ(R) , k);
}
return L;
}
}using namespace Persistent_tree;
void buildtree(int u , int v){
// printf("a[%d] = %d\n" , u , a[u]);
T[u] = update(T[v] , a[u] , 1);
REP(i , SZ(edge[u])){
int go = edge[u][i];
if (go == v) continue;
buildtree(go , u);
}
}
int Q;
void solve(){
REP_1(i , n){
edge[i].clear();
RD(a[i]);
}
hashInit();
REP_1(i , n) a[i] = hashQ(a[i]);
REP_1(i , n - 1){
int x , y;
RD(x) ; RD(y);
edge[x].PB(y);
edge[y].PB(x);
}
initLCA();
dfs(1 , 1 , 0);
make_parent();
tot = 0;
T[0] = build(1 , m);
buildtree(1 , 0);
while(Q--){
int L , R , k , rt;
RD(L) , RD(R) , RD(k);
rt = lca(L , R);
// if (L == R) printf("%d\n" , oQ(a[L]));
// printf("%d %d %d\n" , L , R , rt);
printf("%d\n" , oQ(query(L , R , rt , k)));
}
}
int main(){
// freopen("COT.txt" , "r" , stdin);
while(~scanf("%d%d" , &n , &Q)) solve();
}
D CF 226E —— 记得是当年遗留下来的问题
E BZOJ 1901 ZOJ 2112
ZOJ 2112 是不是得写成动态分配空间啊T_T。不会- -。被卡空间卡死了T_T不知道cxlove 怎么通过的。。也许ZOJ 和 BZOJ 的数据大相径庭？
BZOJ 倒是AC了
题意：
给一个N序列（N<=5e4），Q操作（Q<=1e4）：（1、把某个数改变成另一个数 ； 2、求某段区间内的第k大树）
解法：
首先读入N的序列和Q个询问，离线+hash + 建立主席树组。将主席数组用树状数组的方式规划（算不算树套树？）。可是这样一来空间复杂度就会变成了Const * (NlogN + MlogNlogn)于是乎ZOJ 上面一直在RE 和 MLE 上徘徊。
int n , k;
const int N = 5e4 + 5;
const int hashN = 6e4 + 5;
int a[N];
//namespace HASH{
int h_[hashN] , m;
void hashInitpre(){
for(int i = 1 ; i <= n ; ++i){
h_[i] = a[i];
}
m = n;
}
void hashInitfinal(){
sort(h_ + 1 , h_ + m + 1);
m = unique(h_ + 1 , h_ + m + 1) - h_ - 1;
}
int hashQ(int x){
return lower_bound(h_ + 1, h_ + m + 1 , x) - h_;
}
int oQ(int x){
return h_[x];
}
//}using namespace HASH;
//namespace Persistent_Tree{
const int TN = 2705005;
int T[N];
int l[TN] , r[TN] ;
short c[TN];
int tot;
int build(int L , int R){
int rt = ++ tot;
if (L >= R) return rt;
int mid = L + R >> 1;
l[rt] = build(L , mid);
r[rt] = build(mid + 1 , R);
c[rt] = 0;
return rt;
}
int update(int rt , int p , int d){
int nrt = ++tot , root = nrt;
c[nrt] = c[rt] + d;
int L = 1 , R = m;
int mid;
while(L < R){
mid = L + R >> 1;
if (p <= mid){
l[nrt] = ++tot , r[nrt] = r[rt];
nrt = l[nrt] , rt = l[rt] , R = mid;
}
else{
l[nrt] = l[rt] , r[nrt] = ++tot;
nrt = r[nrt] , rt= r[rt] , L = mid + 1;
}
c[nrt] = c[rt] + d;
}
return root;
}
//}using namespace Persistent_Tree;
//namespace BIT{
int bit[hashN];
void BITinit(int rt , int tn){
tot = 0;
T[rt] = build(1 , tn);
for (int i = 1 ; i <= tn ; ++i)
T[i] = T[rt];
}
void add(int x , int pos , int c){
for ( ; x <= m ; x += low_bit(x))
T[x] = update(T[x] , pos , c);
}
int sum(int x){
int ret = 0;
for( ; x ; x -= low_bit(x))
ret += c[l[bit[x]]];
return ret;
}
int query(int ql , int qr , int k){
for (int i = ql ; i ; i -= low_bit(i))
bit[i] = T[i];
for (int i = qr ; i ; i -= low_bit(i))
bit[i] = T[i];
int L = 1 , R = m;
int mid;
int t;
while(L < R){
mid = L + R >> 1;
t = sum(qr) - sum(ql);
if (t >= k){
for (int i = ql ; i ; i -= low_bit(i))
bit[i] = l[bit[i]];
for (int i = qr ; i ; i -= low_bit(i))
bit[i] = l[bit[i]];
R = mid;
}
else{
k -= t;
for (int i = ql ; i ; i -= low_bit(i))
bit[i] = r[bit[i]];
for (int i = qr ; i ; i -= low_bit(i))
bit[i] = r[bit[i]];
L = mid + 1;
}
// printf("[%2d , %2d] %d\n" , oQ(L) , oQ(R) , k);
}
return L;
}
//}using namespace BIT;
char op[2];
int Q;
const int QN = 1e4 + 5;
struct IO{
bool change;
int x , y , k;
void input(){
RS(op);
if(op[0] == 'C'){
change = 1;
RD(x , y);
h_[ ++m ] = y;
}
else{
change = 0;
RD(x , y , k);
}
}
}q[QN];
const int BZOJ = 0;
void solve(){
if(!BZOJ)RD(n , Q);
REP_1(i , n) RD(a[i]);
hashInitpre();
REP(i , Q)
q[i].input();
hashInitfinal();
REP_1(i , n) a[i] = hashQ(a[i]);
REP(i , Q)
if (q[i].change)
q[i].y = hashQ(q[i].y);
BITinit(0 , m);
REP_1(i , n)
add(i , a[i] , 1);
REP(qq , Q){
if (q[qq].change){
int pos = q[qq].x;
add(pos , a[pos] , -1);
add(pos , q[qq].y , 1);
a[pos] = q[qq].y;
}
else{
OT(oQ(query(q[qq].x - 1 , q[qq].y , q[qq].k)));
}
}
}
int main(){
if (BZOJ)while(~scanf("%d%d" , &n , &Q)) solve();
else{
Rush solve();
}
}
F UVA 12345
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3767
http://user.qzone.qq.com/251815992/blog/1346399378
资料：http://www.shuizilong.com/house/archives/uva-12345-dynamic-lensetalr/
sad...终于搞掉了。。小岛写的还是很NB的。。
题意：
对于一个长度为N （<= 5e4） 的序列，支持一下两种操作：
1、修改某个值
2、求区间第k大
解法：
树状数组 套 主席树 1.084s
这题对树状数组分割区间的理解有很大的帮助啊！！！
还有不要瞎Hash 啊，Hash 一下我就超时了T_T
方法是用个Set记录每个数的pre ， 插入主席树中。。这题相当Nice需要继续参透
#define LOCAL
/** ` Micro Mezzo Macro Flation -- Overheated Economy ., Ver 0.1 **/
#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
#define REP(i, n__) for (int i=0;i<int(n__);++i)
#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
#define REP_1(i, n__) for (int i=1;i<=int(n__);++i)
#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
#define REP_C(i, n) for (int n____=int(n),i=0;i<n____;++i)
#define FOR_C(i, a, b) for (int b____=int(b),i=a;i<b____;++i)
#define DWN_C(i, b, a) for (int a____=int(a),i=b-1;i>=a____;--i)
#define REP_N(i, n) for (i=0;i<int(n);++i)
#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
#define REP_1_C(i, n) for (int n____=int(n),i=1;i<=n____;++i)
#define FOR_1_C(i, a, b) for (int b____=int(b),i=a;i<=b____;++i)
#define DWN_1_C(i, b, a) for (int a____=int(a),i=b;i>=a____;--i)
#define REP_1_N(i, n) for (i=1;i<=int(n);++i)
#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)
#define REP_C_N(i, n) for (n____=int(n),i=0;i<n____;++i)
#define FOR_C_N(i, a, b) for (b____=int(b),i=a;i<b____;++i)
#define DWN_C_N(i, b, a) for (a____=int(a),i=b-1;i>=a____;--i)
#define REP_1_C_N(i, n) for (n____=int(n),i=1;i<=n____;++i)
#define FOR_1_C_N(i, a, b) for (b____=int(b),i=a;i<=b____;++i)
#define DWN_1_C_N(i, b, a) for (a____=int(a),i=b;i>=a____;--i)
//#define ECH(it, A) for (typeof(A.begin()) it=A.begin(); it != A.end(); ++it)
#define ECH(it, A) for (__typeof(A.begin()) it=A.begin(); it != A.end(); ++it)
#define REP_S(it, str) for (char*it=str;*it;++it)
#define REP_G(it, u) for (int it=hd[u];it;it=suc[it])
#define DO(n) for ( int ____n ## __line__ = n; ____n ## __line__ -- ; )
#define REP_2(i, j, n, m) REP(i, n) REP(j, m)
#define REP_2_1(i, j, n, m) REP_1(i, n) REP_1(j, m)
#define REP_3(i, j, k, n, m, l) REP(i, n) REP(j, m) REP(k, l)
#define REP_3_1(i, j, k, n, m, l) REP_1(i, n) REP_1(j, m) REP_1(k, l)
#define ALL(A) A.begin(), A.end()
#define LLA(A) A.rbegin(), A.rend()
#define CPY(A, B) memcpy(A, B, sizeof(A))
#define INS(A, P, B) A.insert(A.begin() + P, B)
#define ERS(A, P) A.erase(A.begin() + P)
#define BSC(A, X) find(ALL(A), X) // != A.end()
#define CTN(T, x) (T.find(x) != T.end())
#define SZ(A) int(A.size())
#define PB push_back
#define MP(A, B) make_pair(A, B)
#define PTT pair<T, T>
#define fi first
#define se second
#define Rush for(int ____T=RD(); ____T--;)
#define Display(A, n, m) { \
REP(i, n){ \
REP(j, m) cout << A[i][j] << " "; \
cout << endl; \
} \
}
#define Display_1(A, n, m) { \
REP_1(i, n){ \
REP_1(j, m) cout << A[i][j] << " "; \
cout << endl; \
} \
}
#pragma comment(linker, "/STACK:36777216")
//#pragma GCC optimize ("O2")
#define Ruby system("ruby main.rb")
#define Haskell system("runghc main.hs")
#define Python system("python main.py")
#define Pascal system("fpc main.pas")
typedef long long LL;
//typedef long double DB;
typedef double DB;
typedef unsigned UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<char> VC;
typedef vector<string> VS;
typedef vector<LL> VL;
typedef vector<DB> VD;
typedef set<int> SI;
typedef set<string> SS;
typedef set<LL> SL;
typedef set<DB> SD;
typedef map<int, int> MII;
typedef map<string, int> MSI;
typedef map<LL, int> MLI;
typedef map<DB, int> MDI;
typedef pair<int, int> PII;
typedef pair<int, bool> PIB;
typedef pair<LL, LL> PLL;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef vector<VII> VVII;
template<class T> inline T& RD(T &);
template<class T> inline void OT(const T &);
inline LL RD(){LL x; return RD(x);}
inline char& RC(char &c){scanf(" %c", &c); return c;}
inline char RC(){char c; return RC(c);}
//inline char& RC(char &c){c = getchar(); return c;}
//inline char RC(){return getchar();}
inline DB& RF(DB &x){scanf("%lf", &x); return x;}
inline DB RF(){DB x; return RF(x);}
inline char* RS(char *s){scanf("%s", s); return s;}
template<class T0, class T1> inline T0& RD(T0 &x0, T1 &x1){RD(x0), RD(x1); return x0;}
template<class T0, class T1, class T2> inline T0& RD(T0 &x0, T1 &x1, T2 &x2){RD(x0), RD(x1), RD(x2); return x0;}
template<class T0, class T1, class T2, class T3> inline T0& RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3){RD(x0), RD(x1), RD(x2), RD(x3); return x0;}
template<class T0, class T1, class T2, class T3, class T4> inline T0& RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4){RD(x0), RD(x1), RD(x2), RD(x3), RD(x4); return x0;}
template<class T0, class T1, class T2, class T3, class T4, class T5> inline T0& RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4, T5 &x5){RD(x0), RD(x1), RD(x2), RD(x3), RD(x4), RD(x5); return x0;}
template<class T0, class T1, class T2, class T3, class T4, class T5, class T6> inline T0& RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4, T5 &x5, T6 &x6){RD(x0), RD(x1), RD(x2), RD(x3), RD(x4), RD(x5), RD(x6); return x0;}
template<class T0, class T1> inline void OT(const T0 &x0, const T1 &x1){OT(x0), OT(x1);}
template<class T0, class T1, class T2> inline void OT(const T0 &x0, const T1 &x1, const T2 &x2){OT(x0), OT(x1), OT(x2);}
template<class T0, class T1, class T2, class T3> inline void OT(const T0 &x0, const T1 &x1, const T2 &x2, const T3 &x3){OT(x0), OT(x1), OT(x2), OT(x3);}
template<class T0, class T1, class T2, class T3, class T4> inline void OT(const T0 &x0, const T1 &x1, const T2 &x2, const T3 &x3, const T4 &x4){OT(x0), OT(x1), OT(x2), OT(x3), OT(x4);}
template<class T0, class T1, class T2, class T3, class T4, class T5> inline void OT(const T0 &x0, const T1 &x1, const T2 &x2, const T3 &x3, const T4 &x4, const T5 &x5){OT(x0), OT(x1), OT(x2), OT(x3), OT(x4), OT(x5);}
template<class T0, class T1, class T2, class T3, class T4, class T5, class T6> inline void OT(const T0 &x0, const T1 &x1, const T2 &x2, const T3 &x3, const T4 &x4, const T5 &x5, const T6 &x6){OT(x0), OT(x1), OT(x2), OT(x3), OT(x4), OT(x5), OT(x6);}
inline char& RC(char &a, char &b){RC(a), RC(b); return a;}
inline char& RC(char &a, char &b, char &c){RC(a), RC(b), RC(c); return a;}
inline char& RC(char &a, char &b, char &c, char &d){RC(a), RC(b), RC(c), RC(d); return a;}
inline char& RC(char &a, char &b, char &c, char &d, char &e){RC(a), RC(b), RC(c), RC(d), RC(e); return a;}
inline char& RC(char &a, char &b, char &c, char &d, char &e, char &f){RC(a), RC(b), RC(c), RC(d), RC(e), RC(f); return a;}
inline char& RC(char &a, char &b, char &c, char &d, char &e, char &f, char &g){RC(a), RC(b), RC(c), RC(d), RC(e), RC(f), RC(g); return a;}
inline DB& RF(DB &a, DB &b){RF(a), RF(b); return a;}
inline DB& RF(DB &a, DB &b, DB &c){RF(a), RF(b), RF(c); return a;}
inline DB& RF(DB &a, DB &b, DB &c, DB &d){RF(a), RF(b), RF(c), RF(d); return a;}
inline DB& RF(DB &a, DB &b, DB &c, DB &d, DB &e){RF(a), RF(b), RF(c), RF(d), RF(e); return a;}
inline DB& RF(DB &a, DB &b, DB &c, DB &d, DB &e, DB &f){RF(a), RF(b), RF(c), RF(d), RF(e), RF(f); return a;}
inline DB& RF(DB &a, DB &b, DB &c, DB &d, DB &e, DB &f, DB &g){RF(a), RF(b), RF(c), RF(d), RF(e), RF(f), RF(g); return a;}
inline void RS(char *s1, char *s2){RS(s1), RS(s2);}
inline void RS(char *s1, char *s2, char *s3){RS(s1), RS(s2), RS(s3);}
template<class T> inline void RST(T &A){memset(A, 0, sizeof(A));}
template<class T0, class T1> inline void RST(T0 &A0, T1 &A1){RST(A0), RST(A1);}
template<class T0, class T1, class T2> inline void RST(T0 &A0, T1 &A1, T2 &A2){RST(A0), RST(A1), RST(A2);}
template<class T0, class T1, class T2, class T3> inline void RST(T0 &A0, T1 &A1, T2 &A2, T3 &A3){RST(A0), RST(A1), RST(A2), RST(A3);}
template<class T0, class T1, class T2, class T3, class T4> inline void RST(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4){RST(A0), RST(A1), RST(A2), RST(A3), RST(A4);}
template<class T0, class T1, class T2, class T3, class T4, class T5> inline void RST(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5){RST(A0), RST(A1), RST(A2), RST(A3), RST(A4), RST(A5);}
template<class T0, class T1, class T2, class T3, class T4, class T5, class T6> inline void RST(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, T6 &A6){RST(A0), RST(A1), RST(A2), RST(A3), RST(A4), RST(A5), RST(A6);}
template<class T> inline void FLC(T &A, int x){memset(A, x, sizeof(A));}
template<class T0, class T1> inline void FLC(T0 &A0, T1 &A1, int x){FLC(A0, x), FLC(A1, x);}
template<class T0, class T1, class T2> inline void FLC(T0 &A0, T1 &A1, T2 &A2, int x){FLC(A0, x), FLC(A1, x), FLC(A2, x);}
template<class T0, class T1, class T2, class T3> inline void FLC(T0 &A0, T1 &A1, T2 &A2, T3 &A3, int x){FLC(A0, x), FLC(A1, x), FLC(A2, x), FLC(A3, x);}
template<class T0, class T1, class T2, class T3, class T4> inline void FLC(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, int x){FLC(A0, x), FLC(A1, x), FLC(A2, x), FLC(A3, x), FLC(A4, x);}
template<class T0, class T1, class T2, class T3, class T4, class T5> inline void FLC(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, int x){FLC(A0, x), FLC(A1, x), FLC(A2, x), FLC(A3, x), FLC(A4, x), FLC(A5, x);}
template<class T0, class T1, class T2, class T3, class T4, class T5, class T6> inline void FLC(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, T6 &A6, int x){FLC(A0, x), FLC(A1, x), FLC(A2, x), FLC(A3, x), FLC(A4, x), FLC(A5, x), FLC(A6, x);}
template<class T> inline void CLR(priority_queue<T, vector<T>, less<T> > &Q){while (!Q.empty()) Q.pop();}
template<class T> inline void CLR(priority_queue<T, vector<T>, greater<T> > &Q){while (!Q.empty()) Q.pop();}
template<class T> inline void CLR(T &A){A.clear();}
template<class T0, class T1> inline void CLR(T0 &A0, T1 &A1){CLR(A0), CLR(A1);}
template<class T0, class T1, class T2> inline void CLR(T0 &A0, T1 &A1, T2 &A2){CLR(A0), CLR(A1), CLR(A2);}
template<class T0, class T1, class T2, class T3> inline void CLR(T0 &A0, T1 &A1, T2 &A2, T3 &A3){CLR(A0), CLR(A1), CLR(A2), CLR(A3);}
template<class T0, class T1, class T2, class T3, class T4> inline void CLR(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4){CLR(A0), CLR(A1), CLR(A2), CLR(A3), CLR(A4);}
template<class T0, class T1, class T2, class T3, class T4, class T5> inline void CLR(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5){CLR(A0), CLR(A1), CLR(A2), CLR(A3), CLR(A4), CLR(A5);}
template<class T0, class T1, class T2, class T3, class T4, class T5, class T6> inline void CLR(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, T6 &A6){CLR(A0), CLR(A1), CLR(A2), CLR(A3), CLR(A4), CLR(A5), CLR(A6);}
template<class T> inline void CLR(T &A, int n){REP(i, n) CLR(A[i]);}
template<class T> inline T& SRT(T &A){sort(ALL(A)); return A;}
template<class T, class C> inline T& SRT(T &A, C B){sort(ALL(A), B); return A;}
// <<= ` 0. Daily Use .,
template<class T> inline void checkMin(T &a,const T b){if (b<a) a=b;}
template<class T> inline void checkMax(T &a,const T b){if (a<b) a=b;}
template<class T> inline void checkMin(T &a, T &b, const T x){checkMin(a, x), checkMin(b, x);}
template<class T> inline void checkMax(T &a, T &b, const T x){checkMax(a, x), checkMax(b, x);}
template <class T, class C> inline void checkMin(T& a, const T b, C c){if (c(b,a)) a = b;}
template <class T, class C> inline void checkMax(T& a, const T b, C c){if (c(a,b)) a = b;}
template<class T> inline T min(T a, T b, T c){return min(min(a, b), c);}
template<class T> inline T max(T a, T b, T c){return max(max(a, b), c);}
template<class T> inline T min(T a, T b, T c, T d){return min(min(a, b), min(c, d));}
template<class T> inline T max(T a, T b, T c, T d){return max(max(a, b), max(c, d));}
template<class T> inline T sqr(T a){return a*a;}
template<class T> inline T cub(T a){return a*a*a;}
inline int Ceil(int x, int y){return (x - 1) / y + 1;}
// <<= ` 1. Bitwise Operation .,
namespace BO{
inline bool _1(int x, int i){return bool(x&1<<i);}
inline bool _1(LL x, int i){return bool(x&1LL<<i);}
inline LL _1(int i){return 1LL<<i;}
inline LL _U(int i){return _1(i) - 1;};
inline int reverse_bits(int x){
x = ((x >> 1) & 0x55555555) | ((x << 1) & 0xaaaaaaaa);
x = ((x >> 2) & 0x33333333) | ((x << 2) & 0xcccccccc);
x = ((x >> 4) & 0x0f0f0f0f) | ((x << 4) & 0xf0f0f0f0);
x = ((x >> 8) & 0x00ff00ff) | ((x << 8) & 0xff00ff00);
x = ((x >>16) & 0x0000ffff) | ((x <<16) & 0xffff0000);
return x;
}
inline LL reverse_bits(LL x){
x = ((x >> 1) & 0x5555555555555555LL) | ((x << 1) & 0xaaaaaaaaaaaaaaaaLL);
x = ((x >> 2) & 0x3333333333333333LL) | ((x << 2) & 0xccccccccccccccccLL);
x = ((x >> 4) & 0x0f0f0f0f0f0f0f0fLL) | ((x << 4) & 0xf0f0f0f0f0f0f0f0LL);
x = ((x >> 8) & 0x00ff00ff00ff00ffLL) | ((x << 8) & 0xff00ff00ff00ff00LL);
x = ((x >>16) & 0x0000ffff0000ffffLL) | ((x <<16) & 0xffff0000ffff0000LL);
x = ((x >>32) & 0x00000000ffffffffLL) | ((x <<32) & 0xffffffff00000000LL);
return x;
}
template<class T> inline bool odd(T x){return x&1;}
template<class T> inline T low_bit(T x) {return x & -x;}
template<class T> inline T high_bit(T x) {T p = low_bit(x);while (p != x) x -= p, p = low_bit(x);return p;}
template<class T> inline T cover_bit(T x){T p = 1; while (p < x) p <<= 1;return p;}
inline int low_idx(int x){return __builtin_ffs(x);}
inline int low_idx(LL x){return __builtin_ffsll(x);}
inline int high_idx(int x){return low_idx(reverse_bits(x));}
inline int high_idx(LL x){return low_idx(reverse_bits(x));}
inline int clz(int x){return __builtin_clz(x);}
inline int clz(LL x){return __builtin_clzll(x);}
inline int ctz(int x){return __builtin_ctz(x);}
inline int ctz(LL x){return __builtin_ctzll(x);}
inline int parity(int x){return __builtin_parity(x);}
inline int parity(LL x){return __builtin_parityll(x);}
inline int lg2(int a){return 31 - __builtin_clz(a);}
inline int count_bits(int x){return __builtin_popcount(x);}
inline int count_bits(LL x){return __builtin_popcountll(x);}
} using namespace BO;
template<class T> inline T& RD(T &x){
//cin >> x;
// scanf("%d", &x);
char c; for (c = getchar(); c < '0'; c = getchar()); x = c - '0'; for (c = getchar(); '0' <= c && c <= '9'; c = getchar()) x = x * 10 + c - '0';
//char c; c = getchar(); x = c - '0'; for (c = getchar(); c >= '0'; c = getchar()) x = x * 10 + c - '0';
return x;
}
int ____Case; template<class T> inline void OT(const T &x){
//if (x == -1) printf("Case %d: NO\n", ++____Case);
//else printf("Case %d: %d\n", ++____Case, x);
//printf("%I64d\n", x);
//printf("%.2lf\n", x);
printf("%d\n", x);
// cout << x << endl;
}
/* .................................................................................................................................. */
const int N = 5e4 + 19;
int a[N] , n , Q;
const int M = 1e6 + 9;
int qx , qy;
namespace Persistent_tree{
const int TN = N * 60 * 50;
int T[N];
int l[TN] , r[TN] , c[TN] , tot , pre[N];
int build(int L , int R){
int rt = ++ tot;
if (L >= R) return rt;
int mid = L + R >> 1;
l[rt] = build(L , mid);
r[rt] = build(mid + 1 , R);
c[rt] = 0;
return rt;
}
int update(int rt , int p , int d){
int nrt = ++tot , root = nrt;
c[nrt] = c[rt] + d;
int L = 0 , R = n;
int mid ;
while (L < R){
mid = L + R >> 1;
if (p <= mid){
l[nrt] = ++ tot , r[nrt] = r[rt];
nrt = l[nrt] , rt = l[rt] , R = mid;
}
else{
l[nrt] = l[rt] , r[nrt] = ++tot;
nrt = r[nrt] , rt = r[rt] , L = mid + 1;
}
c[nrt] = c[rt] + d;
}
return root;
}
int query(int rt){ //Query prefix
int res = 0 , L = 0 , R = n;
int mid ;
while(L < R){
mid = L + R >> 1;
if (qx <= mid){
rt = l[rt];
R = mid;
}
else{
res += c[l[rt]];
rt = r[rt];
L = mid + 1;
}
}
return res;// + c[rt];
}
};
#define PT Persistent_tree
namespace BIT{
void Modify(int rt , int &x , int y){
for ( ; rt <= n ; rt += low_bit(rt)){
PT::T[rt] = PT::update(PT::T[rt] , x , -1) ;
PT::T[rt] = PT::update(PT::T[rt] , y , 1);
}
x = y;
}
int lsum(int rt){
int ret = PT::query(PT::pre[rt]);
for( ; rt ; rt ^= low_bit(rt))
ret += PT::query(PT::T[rt]);
return ret;
}
int query(int l, int r){
return lsum(r) - lsum(l - 1);
}
};
int Pre[N];
SI H[M];
SI :: iterator iter;
int main(){
//freopen("0.txt" , "r" , stdin);
RD(n , Q);
PT::tot = 0;
#define BT BIT
REP_1(i , n)
RD(a[i]);
int Null = PT::build(0 , n);
REP_1(i , n) PT::T[i] = Null;
REP_1(i , n){
iter = H[a[i]].lower_bound(i);
if(iter != H[a[i]].begin())Pre[i] = *--iter;
H[a[i]].insert(i);
}
PT::pre[0] = Null;
REP_1(i , n)
PT::pre[i] = PT::update(PT::pre[i - 1] , Pre[i] , 1);
char op;
DO(Q){
RC(op);
qx = RD() + 1;
qy = RD();
if (op == 'M'){
if (a[qx] == qy) continue;
SI &sx = H[a[qx]] , &sy = H[qy]; sx.erase(qx);
iter = sx.lower_bound(qx);
if(iter != sx.end())
BT::Modify(*iter , Pre[*iter] , Pre[qx]);
iter = sy.lower_bound(qx);
if (iter == sy.begin())
BT::Modify(qx , Pre[qx] , 0);
else
BT::Modify(qx , Pre[qx] , *--sy.lower_bound(qx));
if (iter != sy.end()) BT::Modify(*iter , Pre[*iter] , qx);
sy.insert(qx);
a[qx] = qy;
}
else OT(BT::query(qx , qy));
}
}
G Acdream群第四次群赛C 题
http://www.acdream.net/problem.php?cid=1011&pid=2
资料：http://www.shuizilong.com/house/archives/%E3%80%90%E7%AC%AC%E5%9B%9B%E6%AC%A1-acdream-%E7%BE%A4%E5%8E%9F%E5%88%9B%E7%BE%A4%E8%B5%9B%E3%80%91%E8%A7%A3%E9%A2%98%E6%8A%A5%E5%91%8A/
H CF 140 Div1 D题
资料：http://www.shuizilong.com/house/archives/codeforces-round-140/
I BZOJ 2653
J SPOJ TTM To The Moon
/*
Time 的优化： 每次update的时候更新每个节点的UPD，如果下次更新的时候发现 UPD = Time 也就是这个节点属于本子树，那么直接更新否则加点。
*/
const int N = 1e5 + 1009;
int Time;
struct Node{
Node * lson , *rson;
LL add , sum;
int l , r;
int UPD;
Node(){}
Node(int updtime = Time){
add = 0 , sum = 0;
lson = NULL ; rson = NULL;
UPD = updtime;
}
void pushup(){
sum = 0;
if (lson) sum += lson -> sum;
if (rson) sum += rson -> sum;
}
void pushdown(int updtime);
void copy(Node * pre){
lson = pre -> lson;
rson = pre -> rson;
l = pre -> l;
r = pre -> r;
}
}*C , *T[N];
int cnt;
Node * newNode(int updtime = Time){
C = new Node(updtime);
return C;
}
void Node::pushdown(int updtime){
if (lson && lson -> UPD != updtime){
Node * newlson = newNode(updtime);
newlson -> sum = lson -> sum;
newlson -> add = lson -> add;
newlson -> copy(lson);
lson = newlson;
}
if (rson && rson -> UPD != updtime){
Node * newrson = newNode(updtime);
newrson -> sum = rson -> sum;
newrson -> add = rson -> add;
newrson -> copy(rson);
rson = newrson;
}
if (add == 0) return;
int mid = l + r >> 1;
if (lson){
lson -> sum = lson -> sum + add * (mid - l + 1);
lson -> add = lson -> add + add;
}
if (rson){
rson -> sum = rson -> sum + add * (r - mid);
rson -> add = rson -> add + add;
}
add = 0;
}
void build(Node *& rt , int l , int r){
rt = newNode();
rt -> l = l , rt -> r = r;
if (l == r){
cin >> rt -> sum;
return;
}
int m = l + r >> 1;
if (l <= m) build(rt -> lson , l , m);
if (m < r) build(rt -> rson , m + 1 , r);
rt -> pushup();
}
void update(Node * pre , Node *&rt , int L , int R , int l , int r , LL c , int _time = Time){
if (rt == NULL || rt -> UPD != _time){
rt = newNode(_time);
rt -> add = pre -> add;
rt -> sum = pre -> sum;
rt -> copy(pre);
}
if (L <= l && r <= R){
rt -> add += c;
rt -> sum += c * (r - l + 1);
return;
}
rt -> pushdown(_time);
int m = l + r >> 1;
if (L <= m) update(pre -> lson , rt -> lson , L , R , l , m , c , _time);
if (m < R) update(pre -> rson , rt -> rson , L , R , m + 1 , r , c , _time);
rt -> pushup();
}
LL query(Node * rt , int L , int R , int l , int r , int _time = Time){
if (L <= l && r <= R)
return rt -> sum;
rt -> pushdown(_time);
int m = l + r >> 1;
LL res = 0;
if (L <= m) res += query(rt -> lson , L , R , l , m , _time);
if (m < R) res += query(rt -> rson , L , R , m + 1 , r , _time);
rt -> pushup();
return res;
}
int n , m;
char op[15];
int l , r , t;
LL d;
void work(){
cin >> op;
if (op[0] == 'C'){
T[Time + 1] = NULL;
cin >> l >> r >> d;
update(T[Time] , T[Time + 1] , l , r , 1 , n , d , Time + 1);
Time++;
}
else if (op[0] == 'Q'){
cin >> l >> r;
cout << query(T[Time] , l , r , 1 , n , Time) << endl;
}
else if (op[0] == 'H'){
cin >> l >> r >> t;
cout << query(T[t] , l , r , 1 , n , t) << endl;
}
else{
cin >> Time;
}
}
void solve(){
T[0] = NULL;
C = NULL;
cnt = 0;
Time = 0;
build(T[0] , 1 , n);
DO(m) work();
}
int main(){
ios::sync_with_stdio(false);
while(cin >> n >> m) solve();
}