题解
2024-05-01 19:20:24
发布于:广东
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懒得优化
#include <iostream>
#include <cstdio>
using namespace std;
struct node{
int ct, last;
}a[105][105];
int main(){
int n, m, t1, t2, x1, y1, x2, y2, x, y;
cin >> n >> m >> t1 >> t2;
for(int i = 1; i <= t1; i++){
cin >> x1 >> y1 >> x2 >> y2;
for(int j = x1; j <= x2; j++){
for(int k = y1; k <= y2; k++){
a[j][k].ct++, a[j][k].last = i;
}
}
}while(t2--){
cin >> x >> y;
if(a[x][y].ct){
printf("Y %d %d\n", a[x][y].ct, a[x][y].last);
}else cout << "N\n";
}
return 0;
}
时间复杂度:
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