hdoj 2665 Kth number主席树裸

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9417    Accepted Submission(s): 2938

Problem Description

Give you a sequence and ask you the kth big number of a inteval.

 

Input

The first line is the number of the test cases.
For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere.
The second line contains n integers, describe the sequence.
Each of following m lines contains three integers s, t, k.
[s, t] indicates the interval and k indicates the kth big number in interval [s, t]

 

Output

For each test case, output m lines. Each line contains the kth big number.

 

Sample Input

1 10 1 1 4 2 3 5 6 7 8 9 0 1 3 2

 

Sample Output

2

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#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

const int M = 1000008;

struct Node{
    int r, l, sum;
}T[M*20];
struct A{
    int idx, x;
    bool operator <(const A & o)const{
        return x < o.x;
    }
}a[M];
int T_cnt;
void inser(int &num, int &x, int L, int R){
    T[T_cnt++] = T[x]; x = T_cnt-1;
    ++T[x].sum;
    if(L == R) return;
    int mid = (L+R)>>1;
    if(num <= mid) inser(num, T[x].l, L, mid);
    else inser(num, T[x].r, mid+1, R);
}
int query(int i, int j, int k, int L, int R){
    if(L == R) return L;
    int t = T[T[j].l].sum - T[T[i].l].sum;
    int mid = (R+L)>>1;
    if(k <= t) return query(T[i].l, T[j].l, k, L, mid);
    else return query(T[i].r, T[j].r, k-t, mid+1, R);
}
int ranks[M], root[M];
int n, m, t;
int main(){
    T[0].r = T[0].r = T[0].sum = 0;
    root[0] = 0;
    scanf("%d", &t);
    while(t--){
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i].x);
            a[i].idx = i;
        }
        sort(a+1, a+n+1);
        for(int i = 1; i <= n; i++){
            ranks[a[i].idx] = i;
        }
        T_cnt = 1;
        for(int i = 1; i <= n; i++){
            root[i] = root[i-1];
            inser(ranks[i], root[i], 1, n);
        }
        while(m--){
            int i, j, k;
            scanf("%d%d%d", &i, &j, &k);
            printf("%d
", a[query(root[i-1], root[j], k, 1, n)].x);
        }
    }
}