Yet Another Median Task Gym

题目链接：https://vjudge.net/contest/181019#problem/G

题意：给定一个n*n的矩阵，有q次查询，每次查询给定四个数x1 y1 x2 y2，求矩形区域 x1 <= x <= x1, y1 <= y <= y2的中位数。

思路：若每次对小矩形区域内的所有数进行排序找出中位数，则肯定会超时。STL中有一个函数ntn_element(a, x, a + n)，用于使数组a中n个数中第x大的数位于第x个位置，x左边的数都大于a[x]，x右边的数都小于a[x]但两边的数都可能是无序的。

代码如下：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<cstdlib>#include<sstream>#include<deque>#include<stack>#include<set>#include<map>using namespace std;typedef long long ll;typedef unsigned long long ull;const double eps = 1e-6;const int  maxn = 1000 + 20;const int  maxt = 300 + 10;const int mod = 10;const int dx[] = {1, -1, 0, 0};const int dy[] = {0, 0, -1, 1};const int Dis[] = {-1, 1, -5, 5};const int inf = 0x3f3f3f3f;const int MOD = 1000;int n, m, k;int g[maxn][maxn], num[maxn * maxn];int main(){    int q;    scanf("%d%d", &n, &q);    for(int i = 1; i <= n; ++i){        for(int j = 1; j <= n; ++j){            scanf("%d", &g[i][j]);        }    }    int x1, y1, x2, y2;    while(q--){        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);        int cnt = 0;        for(int i = x1; i <= x2; ++i){            for(int j = y1; j <= y2; ++j){                num[cnt++] = g[i][j];            }        }//        cout << "cnt == " << cnt << " (cnt - 1) / 2 == " << (cnt - 1) / 2 << endl;        nth_element(num, num + (cnt - 1) / 2 ,num + cnt);        printf("%d\n", num[(cnt - 1) / 2]);    }    return 0;}