noi.ac #36 模拟

(des)
存在 (1000 imes 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(n^2logn))

#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
", Answer[x][y]);
	}
	return 0;
}