HDU2665 Kth number 【归并树】

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5425    Accepted Submission(s): 1760

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

 

Source

HDU男生专场公开赛——赶在女生之前先过节（From WHU）

 

Recommend

zty   |   We have carefully selected several similar problems for you:  2660 2662 2667 2663 2661 

跟POJ2104一样。

#include <stdio.h>#include <string.h>#include <algorithm>#include <vector>#define maxn 100005#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1using namespace std;vector<int> T[maxn << 2];int N, Q;void build(int l, int r, int rt) {    if(l == r) {        int val;        scanf("%d", &val);        T[rt].clear();        T[rt].push_back(val);        return;    }    int mid = (l + r) >> 1;    build(lson);    build(rson);    T[rt].resize(r - l + 1); // Attention    merge(T[rt<<1].begin(), T[rt<<1].end(), T[rt<<1|1].begin(), T[rt<<1|1].end(), T[rt].begin());}int query(int L, int R, int val, int l, int r, int rt) {    if(L == l && R == r) {        return upper_bound(T[rt].begin(), T[rt].end(), val) - T[rt].begin();    }    int mid = (l + r) >> 1;    if(R <= mid) return query(L, R, val, lson);    else if(L > mid) return query(L, R, val, rson);    return query(L, mid, val, lson) + query(mid + 1, R, val, rson);}int main() {    int a, b, c, k, left, right, mid, t;    scanf("%d", &t);    while(t--) {        scanf("%d%d", &N, &Q);        build(1, N, 1);        while(Q--) {            scanf("%d%d%d", &a, &b, &k);            left = -1; right = N - 1;            while(right - left > 1) { // binary search                mid = (left + right) >> 1;                c = query(a, b, T[1][mid], 1, N, 1);                if(c >= k) right = mid;                else left = mid;            }            printf("%d\n", T[1][right]);        }    }    return 0;}