树套树（线段树套平衡树）—— ZOJ 2112 Dynamic Rankings
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| Time Limit: 10000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
Description
The Company Dynamic Rankings has developed a new kind of computer that is no longer satisfied with the query like to simply find the k-th smallest number of the given N numbers. They have developed a more powerful system such that for N numbers a[1], a[2], ..., a[N], you can ask it like: what is the k-th smallest number of a[i], a[i+1], ..., a[j]? (For some i<=j, 0<k<=j+1-i that you have given to it). More powerful, you can even change the value of some a[i], and continue to query, all the same.
Your task is to write a program for this computer, which
- Reads N numbers from the input (1 <= N <= 50,000)
- Processes M instructions of the input (1 <= M <= 10,000). These instructions include querying the k-th smallest number of a[i], a[i+1], ..., a[j] and change some a[i] to t.
Input
The first line of the input is a single number X (0 < X <= 4), the number of the test cases of the input. Then X blocks each represent a single test case.
The first line of each block contains two integers N and M, representing N numbers and M instruction. It is followed by N lines. The (i+1)-th line represents the number a[i]. Then M lines that is in the following format
Q i j k or
C i t
It represents to query the k-th number of a[i], a[i+1], ..., a[j] and change some a[i] to t, respectively. It is guaranteed that at any time of the operation. Any number a[i] is a non-negative integer that is less than 1,000,000,000.
There're NO breakline between two continuous test cases.
Output
For each querying operation, output one integer to represent the result. (i.e. the k-th smallest number of a[i], a[i+1],..., a[j])
There're NO breakline between two continuous test cases.
Sample Input
2
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3
Sample Output
3
6
3
6
题意：
n 个数，m 个询问。 Q(l, r, k) 表示询问区间 [l, r] 内第 k 小的数，C(pos, val) 表示把下标为 pos 的数的值该为 val
思路：
树套树入门题，这里选择线段树套Treap树实现。
假设 n 个数用数组 a[] 表示。
线段树可用于区间查询，而平衡树可用于查询区间内某个数的名次；因此可以为线段树的每个区间 [l, r] 都建立一颗以 a[l], a[l+1], ... , a[r] 为键值的平衡树
之后就可以很容易用递归的方法查询 x 在区间 [l, r] 的排名，只需要实现一个 Query(l, r, x)，它返回的是区间 [l, r] 中比 x 小的数的个数，这样 Query(l, r, x) + 1 就是 x 在区间 [l, r] 内的排名；然而题目要求的是求区间 [l, r] 内排名为 k 的数，这怎么办？我们可以用二分答案的办法，具体的过程是：
先找到区间 [l, r] 内的最小值 min 和最大值 max，以此作为二分的两端 beg 和 end（如果嫌麻烦也可以直接使用 a[] 或者取值范围里的最大值最小值）；令 mid_val = (beg + end) / 2，用 Query(l, r, mid_val) 计算 mid_val 的排名，如果 mid_val 排名不大于 k ，则在右半边找，否则在左半边找；二分这里需要注意一些边界问题。
用代码描述就是：
while(beg < end){
mid_val = beg + (end - beg) / 2;
num = Query(l, r, mid_val);
if(num + 1 <= k){
ans = mid_val;
beg = mid + 1;
}
else
end = mid;
}
print(ans);
再下来就是修改某个位置 pos 的值，也就是 C(pos, val) 操作，从树的结构看可以很清晰的想到：需要把线段树中包含 pos 位置的所有区间的平衡树都进行修改。修改的过程不复杂：进入该区间对应的树，找到 a[pos] 并把它删掉，然后在这棵树上插入键值为 val 的结点。
比如要修改 pos = 3 这个位置的值，则需要把图中红色区间对应的每一颗平衡树进行修改。
这样，线段树区间查找的复杂度为 O(logn)，平衡树查找复杂度为 O(logn)，二分查找的复杂度为 O(logn)，则总时间复杂度为 O((logn)^3)，而空间复杂度基本为 lay * n ，lay = logn（即线段树的深度）；此题我使用指针动态分配内存会导致MLE，所以最好使用静态数组。
最后~这道题的 N 要开大一点点。。。
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <string.h>
#define N 70010
int a[N];
int root[N<<2];
int node_count;
typedef struct
{
int val;
int rank;
int sz;
int ch[2];
}Node;
Node node[16*N];
void init()
{
node_count = 0;
memset(root, 0, sizeof(root));
node[0].val = node[0].rank = node[0].sz = node[0].ch[0] = node[0].ch[1] = 0;
}
void push_up(int rt)
{
if(!rt) return;
node[rt].sz = node[node[rt].ch[0]].sz + node[node[rt].ch[1]].sz + 1;
}
void rotate(int *x, int flag)
{
int y = node[*x].ch[flag];
node[*x].ch[flag] = node[y].ch[flag^1];
node[y].ch[flag^1] = *x;
push_up(*x);
(*x) = y;
}
void insert(int *rt, int val)
{
if(!(*rt)){
(*rt) = ++node_count;
node[*rt].val = val;
node[*rt].sz = 1;
node[*rt].rank = rand();
node[*rt].ch[0] = node[*rt].ch[1] = 0;
}
else{
int flag = val > node[*rt].val;
insert(&node[*rt].ch[flag], val);
if(node[node[*rt].ch[flag]].rank < node[*rt].rank)
rotate(rt, flag);
}
push_up(*rt);
}
void build_bst(int *rt, int l, int r)
{
int i;
for(i = l; i <= r; i++)
insert(rt, a[i]);
}
void build(int rt, int left, int right)
{
int mid;
build_bst(&root[rt], left, right);
if(left == right)
return;
mid = left + (right - left) / 2;
build(rt<<1, left, mid);
build(rt<<1|1, mid + 1, right);
}
int small_then_val(int rt, int val)
{
int p = rt;
int ans = 0;
while(p){
if(val > node[p].val){
ans += node[node[p].ch[0]].sz + 1;
p = node[p].ch[1];
}
else
p = node[p].ch[0];
}
return ans;
}
void delete(int *rt, int val)
{
if(!(*rt))
return;
else if(val == node[*rt].val){
if(!node[*rt].ch[0] && !node[*rt].ch[1])
*rt = 0;
else if(!node[*rt].ch[1])
*rt = node[*rt].ch[0];
else if(!node[*rt].ch[0])
*rt = node[*rt].ch[1];
else{
int flag = node[node[*rt].ch[0]].rank > node[node[*rt].ch[1]].rank;
rotate(rt, flag);
delete(&node[*rt].ch[flag^1], val);
}
}
else if(val > node[*rt].val)
delete(&node[*rt].ch[1], val);
else
delete(&node[*rt].ch[0], val);
push_up(*rt);
}
void update(int rt, int left, int right, int pos, int val)
{
int mid;
delete(&root[rt], a[pos]);
insert(&root[rt], val);
if(left == right)
return;
mid = left + (right - left) / 2;
if(pos > mid)
update(rt<<1|1, mid + 1, right, pos, val);
else
update(rt<<1, left, mid, pos, val);
}
int Query(int rt, int left, int right, int l, int r, int val)
{
int mid;
if(left == l && right == r)
return small_then_val(root[rt], val);
mid = left + (right - left) / 2;
if(l > mid)
return Query(rt<<1|1, mid + 1, right, l, r, val);
else if(r <= mid)
return Query(rt<<1, left, mid, l, r, val);
else
return Query(rt<<1, left, mid, l, mid, val) + Query(rt<<1|1, mid + 1, right, mid + 1, r, val);
}
int main()
{
#if 0
freopen("in.txt","r",stdin);
#endif
int T;
srand((unsigned int)time(NULL));
scanf("%d", &T);
while(T--){
int i;
int n, m;
init();
scanf("%d%d", &n, &m);
for(i = 0; i < n; i++)
scanf("%d", a + i);
build(1, 0, n - 1);
for(i = 0; i < m; i++){
int l, r, k, pos, val;
int beg = 0;
int end = 1000000000;
int num, mid, ans;
char od[5];
scanf("%s", od);
if(od[0] == 'Q'){
scanf("%d%d%d", &l, &r, &k);
l--; r--;
while(beg < end){
mid = beg + (end - beg) / 2;
num = Query(1, 0, n - 1, l, r, mid);
if(num + 1 <= k){
ans = mid;
beg = mid + 1;
}
else
end = mid;
}
printf("%d\n", ans);
}
else{
scanf("%d%d", &pos, &val);
pos--;
update(1, 0, n - 1, pos, val);
a[pos] = val;
}
}
}
return 0;
}