noi.ac #36 模拟

\(des\)
存在 \(1000 \times 1000\) 的矩阵，保证元素互不相同，\(2e5\) 次询问，每次询

问给定 \(x, y\) 问存在多少点 \((a, b)\) 满足该元素是 \(a\) 行的 \(x\) 大， \(b\)

列的 \(y\) 大。

\(sol\)

这数据范围给的不敢写暴力啊，然而这 T1 就是暴力啊
需要对所有可能的情况预处理，处理处所有可能的询问
时间复杂度 \(O(nlogn + mlogm)\)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>

using namespace std;
const int N = 1010;

#define gc getchar()
#define Rep(i, a, b) for(int i = a; i <= b; i ++)
#define LL long long

inline int read() {int x = 0; char c = gc; while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x;}
inline LL readLL() {LL x = 0; char c = gc; while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x;}

int A[N][N], B[N][N];
int Sa[N][N], Sb[N][N];
int Answer[N][N];
int n, m, q;

int main() {
    n = read(), m = read(), q = read();
    Rep(i, 1, n) Rep(j, 1, m) A[i][j] = read();
    Rep(i, 1, m) Rep(j, 1, n) B[i][j] = A[j][i];
    Rep(i, 1, n) Rep(j, 1, m) Sa[i][j] = A[i][j];
    Rep(i, 1, m) Rep(j, 1, n) Sb[i][j] = B[i][j];
    Rep(i, 1, n) sort(Sa[i] + 1, Sa[i] + m + 1);
    Rep(i, 1, m) sort(Sb[i] + 1, Sb[i] + n + 1);
    Rep(i, 1, n) 
        Rep(j, 1, m) {
            int Num = A[i][j];
            int x = lower_bound(Sa[i] + 1, Sa[i] + m + 1, Num) - Sa[i];
            int y = lower_bound(Sb[j] + 1, Sb[j] + n + 1, Num) - Sb[j];
            int xx = m - x + 1, yy = n - y + 1;
            Answer[xx][yy] ++;
        }
    Rep(i, 1, q) {
        int x = read(), y = read();
        printf("%d\n", Answer[x][y]);
    }
    return 0;
}