HDU 2665 Kth number

Kth number

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

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

题意：

和poj2104差不多这里就不多写了，这里是题解：查看题解
需要注意的是cnt和vector的初始化。

 #include 
#include 
#include 
#include 
using namespace std;
const int maxn = 1e+5 + 10;
struct NODE {
    int l;
    int r;
    int sum;
};
NODE T[maxn * 50];
int root[maxn], a[maxn];
int cnt = 0;
vector v;
int GetId(int x) {
    return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
}
void Updata(int l, int r, int &x, int y, int pos) {
    T[++cnt] = T[y];
    T[cnt].sum++;
    x = cnt;
    if (l == r) return ;
    int mid = (l + r) >> 1;
    if (mid >= pos) Updata(l, mid, T[x].l, T[y].l, pos);
    else Updata(mid + 1, r, T[x].r, T[y].r, pos);
}
int Query(int l, int r, int x, int y, int k) {
    if (l == r) return l;
    int mid = (l + r) >> 1;
    int sum = T[T[y].l].sum - T[T[x].l].sum;
    if (sum >= k) return Query(l, mid, T[x].l, T[y].l, k);
    else return Query(mid + 1, r, T[x].r, T[y].r, k - sum);
}
int main() {
    int n, t, m;
    scanf("%d", &t);
    while (t--) {
        cnt = 0;
        v.clear();
        scanf("%d %d", &n, &m);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            v.push_back(a[i]);
        }
        sort(v.begin(), v.end());
        v.erase(unique(v.begin(), v.end()), v.end());
        for (int i = 1; i <= n; i++) Updata(1, n, root[i], root[i - 1], GetId(a[i]));
        while (m--) {
            int x, y, k;
            scanf("%d %d %d", &x, &y, &k);
            printf("%d\n", v[Query(1, n, root[x - 1], root[y], k) - 1]);
        }
    }
    return 0;
}