数据结构---各种树模板 持续更新···
1：Trie 字典树
字典树
又称单词查找树，Trie树，是一种树形结构，是一种哈希树的变种。典型应用是用于统计，排序和保存大量的字符串（但不仅限于字符串），所以经常被搜索引擎系统用于文本词频统计。它的优点是：利用字符串的公共前缀来节约存储空间，最大限度地减少无谓的字符串比较，查询效率比哈希表高。
它有3个基本性质：根节点不包含字符，除根节点外每一个节点都只包含一个字符。 从根节点到某一节点，路径上经过的字符连接起来，为该节点对应的字符串。 每个节点的所有子节点包含的字符都不相同。
详细应用见 http://blog.csdn.net/metalseed/article/details/7953262#include <iostream>
using namespace std;
#define MAX 26 //字符集大小
typedef struct TrieNode {
int nCount;
struct TrieNode *next[MAX];
}TrieNode;
TrieNode Memory[1000000];
int allocp =0;
TrieNode *CreateTrieNode() {
int i;
TrieNode *p;
p = &Memory[allocp++];
p->nCount = 1;
for(i =0 ; i < MAX ; i++) {
p->next[i] = NULL;
}
return p;
}
void InsertTrie(TrieNode * &pRoot , char*s) {
int i, k;
TrieNode *p;
if(!(p = pRoot)) {
p = pRoot = CreateTrieNode();
}
i = 0;
while(s[i]) {
k = s[i++] - 'a';
if(p->next[k])
p->next[k]->nCount++;
else
p->next[k] = CreateTrieNode();
p = p->next[k];
}
}
/*查询该前缀出现次数*/
/*若查询该词出现次数
则所有的count初始化为0
并去除所有对count的操作
只在每次InsertTrie末尾加 p->nCount++;
*/
int SearchTrie(TrieNode * &pRoot , char*s) {
TrieNode *p;
int i , k;
if(!(p = pRoot)) {
return 0;
}
i = 0;
while(s[i]) {
k = s[i++] -'a';
if(p->next[k] == NULL) return 0;
p = p->next[k];
}
return p->nCount;
}
int main()
{
TrieNode *ROOT = NULL;
InsertTrie(ROOT,"see");
InsertTrie(ROOT,"seeda");
InsertTrie(ROOT,"seedb");
InsertTrie(ROOT,"seeda");
cout<<SearchTrie(ROOT,"seeda")<<endl;
cout<<SearchTrie(ROOT,"seedb")<<endl;
cout<<SearchTrie(ROOT,"see")<<endl;
return 0;
}
3：BIT 树状数组
1、概述
树状数组(binary indexed tree)，是一种设计新颖的数组结构，它能够高效地获取数组中连续n个数的和。概括说，树状数组通常用于解决以下问题：数组{a}中的元素可能不断地被修改，怎样才能快速地获取连续几个数的和？
2、树状数组基本操作
传统数组(共n个元素)的元素修改和连续元素求和的复杂度分别为O(1)和O(n)。树状数组通过将线性结构转换成伪树状结构（线性结构只能逐个扫描元素，而树状结构可以实现跳跃式扫描），使得修改和求和复杂度均为O(lgn)，大大提高了整体效率。
详见 ： http://blog.csdn.net/metalseed/article/details/7984075 （含2D模板）#include<iostream>
#include<algorithm>
using namespace std;
/* 1D */
const int maxn = 10000;
int tree[maxn],maxID = 0xff/*数组长度，由题定，建树用update*/;
inline void init()
{
memset(tree,0,sizeof(tree));
}
inline int read(int idx){ /*求前idx项和*/
int sum = 0;
while (idx > 0){
sum += tree[idx];
idx -= (idx & -idx);
}
return sum;
}
inline int readSingle(int idx){ /*求idx单点值*/
int sum = tree[idx];
if (idx > 0){
int z = idx - (idx & -idx);
idx--;
while (idx != z){
sum -= tree[idx];
idx -= (idx & -idx);
}
}
return sum;
}
inline void update(int idx ,int val){ /*给idx位置叠加val,叠加！*/
while (idx <= maxID){
tree[idx] += val;
idx += (idx & -idx);
}
}
int main()
{
cout << read(10) << endl;
for(int i = 0; i< 10; ++i)
{
update(i+1,i+1);
}
update(4,999);
cout << read(1) << endl;
cout << read(3) << endl;
cout << readSingle(4) << endl;
return 0;
}
4：Segment Tree 线段树
成段加C 区间求和
#include <cstdio>
#include <algorithm>
using namespace std;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define LL long long
const int maxn = 111111;
/* Node Begin */
LL add[maxn<<2];
LL sum[maxn<<2];
/* Node End */
void PushUp(int rt) {
sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void PushDown(int rt,int m) {
if (add[rt]) {
add[rt<<1] += add[rt];
add[rt<<1|1] += add[rt];
sum[rt<<1] += add[rt] * (m - (m >> 1));
sum[rt<<1|1] += add[rt] * (m >> 1);
add[rt] = 0;
}
}
void build(int l,int r,int rt) {
add[rt] = 0;
if (l == r) {
scanf("%lld",&sum[rt]);
return ;
}
int m = (l + r) >> 1;
build(lson);
build(rson);
PushUp(rt);
}
void update(int L,int R,int c,int l,int r,int rt) {
if (L <= l && r <= R) {
add[rt] += c;
sum[rt] += (LL)c * (r - l + 1);
return ;
}
PushDown(rt , r - l + 1);
int m = (l + r) >> 1;
if (L <= m) update(L , R , c , lson);
if (m < R) update(L , R , c , rson);
PushUp(rt);
}
LL query(int L,int R,int l,int r,int rt) {
if (L <= l && r <= R) {
return sum[rt];
}
PushDown(rt , r - l + 1);
int m = (l + r) >> 1;
LL ret = 0;
if (L <= m) ret += query(L , R , lson);
if (m < R) ret += query(L , R , rson);
return ret;
}
int main() {
int N , Q;
scanf("%d%d",&N,&Q);
build(1 , N , 1);
while (Q --) {
char op[2];
int a , b , c;
scanf("%s",op);
if (op[0] == 'Q')
{
scanf("%d%d",&a,&b);
printf("%lld\n",query(a , b , 1 , N , 1));
}
else
{
scanf("%d%d%d",&a,&b,&c);
update(a , b , c , 1 , N , 1);
}
}
return 0;
}
5：SBT 二叉平衡检索树
http://acm.hdu.edu.cn/showproblem.php?pid=4006
#include <stdio.h>
#include <string.h>
#define MAX 1000010
int n,m;
struct SBT {
int left,right,size,key;
void Init() {
left = right = 0;
size = 1;
}
}a[MAX];
int tot,root;
void left_rotate(int &t) {
int k = a[t].right;
a[t].right = a[k].left;
a[k].left = t;
a[k].size = a[t].size;
a[t].size = a[a[t].left].size + a[a[t].right].size + 1;
t = k;
}
void right_rotate(int &t) {
int k = a[t].left;
a[t].left = a[k].right;
a[k].right = t;
a[k].size = a[t].size;
a[t].size = a[a[t].left].size + a[a[t].right].size + 1;
t = k;
}
void maintain(int &t,int flag) {
if (flag == 0) {
if (a[a[a[t].left].left].size > a[a[t].right].size)
right_rotate(t);
else if (a[a[a[t].left].right].size > a[a[t].right].size)
left_rotate(a[t].left),right_rotate(t);
else return;
}
else {
if (a[a[a[t].right].right].size > a[a[t].left].size)
left_rotate(t);
else if (a[a[a[t].right].left].size > a[a[t].left].size)
right_rotate(a[t].right),left_rotate(t);
else return;
}
maintain(a[t].left,0);
maintain(a[t].right,1);
maintain(t,0);
maintain(t,1);
}
void insert(int &t,int v) {
if (t == 0)
t = ++tot,a[t].Init(),a[t].key = v;
else {
a[t].size++;
if (v < a[t].key)
insert(a[t].left,v);
else insert(a[t].right,v);
maintain(t,v>=a[t].key);
}
}
int del(int &t,int v) {
if (!t) return 0;
a[t].size--;
if (v == a[t].key || v < a[t].key && !a[t].left
|| v > a[t].key && !a[t].right) {
if (a[t].left && a[t].right) {
int p = del(a[t].left,v+1);
a[t].key = a[p].key;
return p;
}
else {
int p = t;
t = a[t].left + a[t].right;
return p;
}
}
else return del(v<a[t].key?a[t].left:a[t].right,v);
}
int find(int t,int k) {
if (k <= a[a[t].left].size)
return find(a[t].left,k);
if (k > a[a[t].left].size + 1)
return find(a[t].right,k-a[a[t].left].size-1);
return a[t].key;
}
int main()
{
int i,j,k;
while (scanf("%d%d",&n,&m) != EOF) {
tot = root = 0;
char ope[10];
while (n--) {
scanf("%s",ope);
if (ope[0] == 'I') {
scanf("%d",&k);
insert(root,k);
}
else printf("%d\n",find(root,a[root].size+1-m));
}
}
}
6：Splay 伸展树
/*
题意：有N个数字围成一个圈，有M个操作，操作类型有六种：
（1）“add x"，从当前指针位置开始的顺时针K2个数加上x。
（2）"reverse"，翻转，从当前指针指针位置开始的顺时针的K2个数。
（3）"insert x"，在当前指针位置的顺时候方向插入一个数x。
（4）”delete“，删除当前指针所指的数。
（5）"move x”，如果x=1，指针逆时针旋转，如果x=2，顺时针旋转。
（6）“query"，查询指针所指向的数的值。
Splay的作法就不说了。
还可以有三个双端队列加两个标记搞定，方法，是第一个双端队列que1维护前K1个数，
第二个que2维护第K1+1到第K2个数，第三个que3维护接下的数，标记add，
表示que1和que2里的数要加上多少，标记head，表示que1是否被翻转过。
Splay（姿势1）：
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
#define LL(x) (ch[x][0])
#define RR(x) (ch[x][1])
#define Kt (ch[ ch[Rt][1] ][0])
#define MID(a,b) (a+((b-a)>>1))
const int N=1e6+5;
int n,m,k1,k2;
int a[N/2];
struct SplayTree
{
int Rt,top;
int pre[N],sz[N],ch[N][2];
int key[N],add[N],pos;
bool flip[N];
inline void Link(int x,int y,int f)
{
pre[x]=y; if(y) ch[y][f]=x;
}
inline void Rotate(int x,int f)
{
int y=pre[x],z=pre[y];
PushDown(y); PushDown(x);
Link(x,z,RR(z)==y);
Link(ch[x][f],y,!f);
Link(y,x,f);
PushUp(y);
}
inline void Splay(int x,int goal)
{
while(pre[x]!=goal)
{
int y=pre[x],z=pre[y];
int cx=(LL(y)==x),cy=(LL(z)==y);
if(z==goal) Rotate(x,cx);
else
{
if(cx==cy) Rotate(y,cy);
else Rotate(x,cx);
Rotate(x,cy);
}
}
PushUp(x);
if(goal==0) Rt=x;
}
inline void Select(int K,int goal)
{
int x=Rt;
PushDown(x);
while(1)
{
if(sz[LL(x)]>=K) x=LL(x);
else if(sz[LL(x)]+1==K) break;
else K-=sz[LL(x)]+1,x=RR(x);
PushDown(x);
}
Splay(x,goal);
}
inline void fun_add(int x,int valu)
{
add[x]+=valu;
key[x]+=valu;
}
inline void fun_flip(int x)
{
flip[x]^=1;
swap(LL(x),RR(x));
}
inline void PushDown(int x)
{
if(add[x])
{
fun_add(LL(x),add[x]);
fun_add(RR(x),add[x]);
add[x]=0;
}
if(flip[x])
{
fun_flip(LL(x));
fun_flip(RR(x));
flip[x]=0;
}
}
inline void PushUp(int x)
{
sz[x]=1+sz[LL(x)]+sz[RR(x)];
}
inline void Add(int x)
{
Select(1,0); Select(k2+2,Rt);
fun_add(Kt,x);
}
inline void Reverse()
{
Select(1,0); Select(k1+2,Rt);
fun_flip(Kt);
}
inline void Insert(int x,int pos)
{
Select(pos,0); Select(pos+1,Rt);
addNode(x,Kt,RR(Rt));
PushUp(RR(Rt)); PushUp(Rt);
}
inline int Delete(bool top)
{
int valu;
if(top)
{
Select(1,0); Select(3,Rt);
valu=key[Kt];
Kt=0;
PushUp(RR(Rt)); PushUp(Rt);
}
else
{
int len=sz[Rt];
Select(len-2,0);Select(len,Rt);
valu=key[Kt];
Kt=0;
PushUp(RR(Rt)); PushUp(Rt);
}
return valu;
}
inline void Move(int x)
{
if(x==1)
{
int valu=Delete(0);
Insert(valu,1);
}
else
{
int valu=Delete(1);
Insert(valu,sz[Rt]-1);
}
}
inline void Query()
{
Select(2,0);
printf("%d\n",key[Rt]);
}
// void Debug(){ printf("Rt:%d\n",Rt); Travel(Rt); }
// void Travel(int x)
// {
// if(x==0) return;
//
// PushDown(x);
// Travel(LL(x));
// printf("node:%d,pre:%d,sz:%d,LL:%d,RR:%d,key:%d\n",
// x,pre[x],sz[x],LL(x),RR(x),key[x]);
// Travel(RR(x));
// }
void addNode(int valu,int &x,int f)
{
x=++top;
pre[x]=f; sz[x]=1; LL(x)=RR(x)=0;
key[x]=valu; add[x]=flip[x]=0;
}
void build(int lft,int rht,int &x,int f)
{
if(lft>rht) return;
int mid=MID(lft,rht);
addNode(a[mid],x,f);
build(lft,mid-1,LL(x),x);
build(mid+1,rht,RR(x),x);
PushUp(x);
}
void init()
{
Rt=top=0;
pre[0]=sz[0]=LL(0)=RR(0)=0;
addNode(0,Rt,0); addNode(0,RR(Rt),Rt);
build(0,n-1,Kt,RR(Rt));
PushUp(RR(Rt)); PushUp(Rt);
}
}spt;
int main()
{
int t_cnt=0;
while(scanf("%d%d%d%d",&n,&m,&k1,&k2)!=EOF)
{
if(n==0&&m==0&&k1==0&&k2==0) break;
for(int i=0;i<n;i++) scanf("%d",&a[i]);
spt.init();
printf("Case #%d:\n",++t_cnt);
char op[100]; int x;
while(m--)
{
scanf("%s",op);
if(op[0]=='a')
{
scanf("%d",&x); spt.Add(x);
}
else if(op[0]=='r') spt.Reverse();
else if(op[0]=='i')
{
scanf("%d",&x); spt.Insert(x,2);
}
else if(op[0]=='d') spt.Delete(1);
else if(op[0]=='m')
{
scanf("%d",&x); spt.Move(x);
}
else spt.Query();
}
}
return 0;
}
Spaly（姿势2）：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define LL(x) (ch[x][0])
#define RR(x) (ch[x][1])
#define Kt (ch[ ch[Rt][1] ][0])
#define MID(a,b) (a+((b-a)>>1))
const int N=1e6+5;
int a[N/2];
int n,m,K1,K2,pos;
struct SplayTree
{
int Rt,top;
int sz[N],pre[N],ch[N][2];
bool flip[N];
int key[N],add[N];
inline void Link(int x,int y,int f)
{
pre[x]=y; if(y) ch[y][f]=x;
}
inline void Rotate(int x,int f)
{
int y=pre[x],z=pre[y];
PushDown(y); PushDown(x);
Link(x,z,RR(z)==y);
Link(ch[x][f],y,!f);
Link(y,x,f);
PushUp(y);
}
inline void Splay(int x,int goal)
{
while(pre[x]!=goal)
{
int y=pre[x],z=pre[y];
int cx=(LL(y)==x),cy=(LL(z)==y);
if(z==goal) Rotate(x,cx);
else
{
if(cx==cy) Rotate(y,cy);
else Rotate(x,cx);
Rotate(x,cy);
}
}
PushUp(x);
if(goal==0) Rt=x;
}
inline void Select(int K,int goal)
{
int x=Rt;
PushDown(x);
while(1)
{
if(sz[LL(x)]>=K) x=LL(x);
else if(1+sz[LL(x)]==K) break;
else K-=sz[LL(x)]+1,x=RR(x);
PushDown(x);
}
Splay(x,goal);
}
inline void fun_add(int x,int valu)
{
key[x]+=valu;
add[x]+=valu;
}
inline void fun_flip(int x)
{
flip[x]^=1;
swap(LL(x),RR(x));
}
inline void PushUp(int x)
{
sz[x]=1+sz[LL(x)]+sz[RR(x)];
}
inline void PushDown(int x)
{
if(add[x])
{
fun_add(LL(x),add[x]);
fun_add(RR(x),add[x]);
add[x]=0;
}
if(flip[x])
{
fun_flip(LL(x)); fun_flip(RR(x));
flip[x]=0;
}
}
inline void Add(int st,int ed,int valu)
{
Select(st-1,0); Select(ed+1,Rt);
fun_add(Kt,valu);
}
inline void Reverse(int st,int ed)
{
Select(st-1,0); Select(ed+1,Rt);
fun_flip(Kt);
}
inline void Insert(int pos,int valu)
{
Select(pos,0); Select(pos+1,Rt);
addNode(valu,Kt,RR(Rt));
PushUp(RR(Rt)); PushUp(Rt);
}
inline void Delete(int pos)
{
Select(pos-1,0); Select(pos+1,Rt);
Kt=0; PushUp(RR(Rt)); PushUp(Rt);
}
inline void Query(int pos)
{
Select(pos,0);
printf("%d\n",key[Rt]);
}
inline void Move(int len)
{
pos-=len;
Select(1,0); Select(2+len,Rt);
int r1=Kt; Kt=0;
PushUp(RR(Rt)); PushUp(Rt);
Select(sz[Rt]-1,0); Select(sz[Rt],Rt);
Link(r1,RR(Rt),0);
PushUp(RR(Rt)); PushUp(Rt);
}
inline void addNode(int valu,int &x,int f)
{
x=++top;
sz[x]=1; pre[x]=f; LL(x)=RR(x)=0;
key[x]=valu; add[x]=flip[x]=0;
}
void build(int lft,int rht,int &x,int f)
{
if(lft>rht) return;
int mid=MID(lft,rht);
addNode(a[mid],x,f);
build(lft,mid-1,LL(x),x);
build(mid+1,rht,RR(x),x);
PushUp(x);
}
void init()
{
Rt=top=0;
addNode(0,Rt,0); addNode(0,RR(Rt),Rt);
build(0,n-1,Kt,RR(Rt));
PushUp(RR(Rt)); PushUp(Rt);
}
// void Debug(){ printf("Rt:%d\n",Rt); Travel(Rt); }
// void Travel(int x)
// {
// if(x==0) return;
// PushDown(x);
// Travel(LL(x));
// printf("node:%d,sz:%d,pre:%d,LL:%d,RR:%d,key:%d\n",
// x,sz[x],pre[x],LL(x),RR(x),key[x]);
// Travel(RR(x));
// }
}spt;
void deal(int &pos,int len)
{
if(pos<=1) pos=len-1;
if(pos>=len) pos=2;
}
int main()
{
freopen("in.txt","r",stdin);
int t_cnt=0;
while(scanf("%d%d%d%d",&n,&m,&K1,&K2)!=EOF)
{
if(n==0&&m==0&&K1==0&&K2==0) break;
for(int i=0;i<n;i++) scanf("%d",&a[i]);
spt.init(); pos=2;
printf("Case #%d:\n",++t_cnt);
char op[100]; int x;
while(m--)
{
int len=spt.sz[spt.Rt];
scanf("%s",op);
if(op[0]=='a')
{
scanf("%d",&x);
if(pos+K2>len) spt.Move(pos+K2-len);
spt.Add(pos,pos+K2-1,x);
}
else if(op[0]=='r')
{
if(pos+K1>=len) spt.Move(pos+K1-len);
spt.Reverse(pos,pos+K1-1);
}
else if(op[0]=='i')
{
scanf("%d",&x); spt.Insert(pos,x);
}
else if(op[0]=='d')
{
spt.Delete(pos);
deal(pos,spt.sz[spt.Rt]);
}
else if(op[0]=='m')
{
scanf("%d",&x);
if(x==1) pos--;
else pos++;
deal(pos,len);
}
else if(op[0]=='q') spt.Query(pos);
// spt.Debug();
}
}
return 0;
}
/*
Description
You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4
Sample Output
4
55
9
15
http://acm.pku.edu.cn/JudgeOnline/problem?id=3468
区间跟新,区间求和
*/
#include <cstdio>
#define keyTree (ch[ ch[root][1] ][0])
const int maxn = 222222;
struct SplayTree{
int sz[maxn];
int ch[maxn][2];
int pre[maxn];
int root , top1 , top2;
int ss[maxn] , que[maxn];
inline void Rotate(int x,int f) {
int y = pre[x];
push_down(y);
push_down(x);
ch[y][!f] = ch[x][f];
pre[ ch[x][f] ] = y;
pre[x] = pre[y];
if(pre[x]) ch[ pre[y] ][ ch[pre[y]][1] == y ] = x;
ch[x][f] = y;
pre[y] = x;
push_up(y);
}
inline void Splay(int x,int goal) {
push_down(x);
while(pre[x] != goal) {
if(pre[pre[x]] == goal) {
Rotate(x , ch[pre[x]][0] == x);
} else {
int y = pre[x] , z = pre[y];
int f = (ch[z][0] == y);
if(ch[y][f] == x) {
Rotate(x , !f) , Rotate(x , f);
} else {
Rotate(y , f) , Rotate(x , f);
}
}
}
push_up(x);
if(goal == 0) root = x;
}
inline void RotateTo(int k,int goal) {//把第k位的数转到goal下边
int x = root;
push_down(x);
while(sz[ ch[x][0] ] != k) {
if(k < sz[ ch[x][0] ]) {
x = ch[x][0];
} else {
k -= (sz[ ch[x][0] ] + 1);
x = ch[x][1];
}
push_down(x);
}
Splay(x,goal);
}
inline void erase(int x) {//把以x为祖先结点删掉放进内存池,回收内存
int father = pre[x];
int head = 0 , tail = 0;
for (que[tail++] = x ; head < tail ; head ++) {
ss[top2 ++] = que[head];
if(ch[ que[head] ][0]) que[tail++] = ch[ que[head] ][0];
if(ch[ que[head] ][1]) que[tail++] = ch[ que[head] ][1];
}
ch[ father ][ ch[father][1] == x ] = 0;
pushup(father);
}
//以上一般不修改//////////////////////////////////////////////////////////////////////////////
void debug() {printf("%d\n",root);Treaval(root);}
void Treaval(int x) {
if(x) {
Treaval(ch[x][0]);
printf("结点%2d:左儿子 %2d 右儿子 %2d 父结点 %2d size = %2d ,val = %2d\n",x,ch[x][0],ch[x][1],pre[x],sz[x],val[x]);
Treaval(ch[x][1]);
}
}
//以上Debug
//以下是题目的特定函数:
inline void NewNode(int &x,int c) {
if (top2) x = ss[--top2];//用栈手动压的内存池
else x = ++top1;
ch[x][0] = ch[x][1] = pre[x] = 0;
sz[x] = 1;
val[x] = sum[x] = c;/*这是题目特定函数*/
add[x] = 0;
}
//把延迟标记推到孩子
inline void push_down(int x) {/*这是题目特定函数*/
if(add[x]) {
val[x] += add[x];
add[ ch[x][0] ] += add[x];
add[ ch[x][1] ] += add[x];
sum[ ch[x][0] ] += (long long)sz[ ch[x][0] ] * add[x];
sum[ ch[x][1] ] += (long long)sz[ ch[x][1] ] * add[x];
add[x] = 0;
}
}
//把孩子状态更新上来
inline void push_up(int x) {
sz[x] = 1 + sz[ ch[x][0] ] + sz[ ch[x][1] ];
/*这是题目特定函数*/
sum[x] = add[x] + val[x] + sum[ ch[x][0] ] + sum[ ch[x][1] ];
}
/*初始化*/
inline void makeTree(int &x,int l,int r,int f) {
if(l > r) return ;
int m = (l + r)>>1;
NewNode(x , num[m]); /*num[m]权值改成题目所需的*/
makeTree(ch[x][0] , l , m - 1 , x);
makeTree(ch[x][1] , m + 1 , r , x);
pre[x] = f;
push_up(x);
}
inline void init(int n) {/*这是题目特定函数*/
ch[0][0] = ch[0][1] = pre[0] = sz[0] = 0;
add[0] = sum[0] = 0;
root = top1 = 0;
//为了方便处理边界,加两个边界顶点
NewNode(root , -1);
NewNode(ch[root][1] , -1);
pre[top1] = root;
sz[root] = 2;
for (int i = 0 ; i < n ; i ++) scanf("%d",&num[i]);
makeTree(keyTree , 0 , n-1 , ch[root][1]);
push_up(ch[root][1]);
push_up(root);
}
/*更新*/
inline void update( ) {/*这是题目特定函数*/
int l , r , c;
scanf("%d%d%d",&l,&r,&c);
RotateTo(l-1,0);
RotateTo(r+1,root);
add[ keyTree ] += c;
sum[ keyTree ] += (long long)c * sz[ keyTree ];
}
/*询问*/
inline void query() {/*这是题目特定函数*/
int l , r;
scanf("%d%d",&l,&r);
RotateTo(l-1 , 0);
RotateTo(r+1 , root);
printf("%lld\n",sum[keyTree]);
}
/*这是题目特定变量*/
int num[maxn];
int val[maxn];
int add[maxn];
long long sum[maxn];
}spt;
int main() {
int n , m;
scanf("%d%d",&n,&m);
spt.init(n);
while(m --) {
char op[2];
scanf("%s",op);
if(op[0] == 'Q') {
spt.query();
} else {
spt.update();
}
}
return 0;
}
7：动态树
8：斯坦纳树
9：主席树
10：QTree