hdu 5480 Conturbatio 区间和

Conturbatio

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 553    Accepted Submission(s): 257

Problem Description

There are many rook on a chessboard, a rook can attack the row and column it belongs, including its own place.

There are also many queries, each query gives a rectangle on the chess board, and asks whether every grid in the rectangle will be attacked by any rook?

 

Input

The first line of the input is a integer T, meaning that there are T test cases.

Every test cases begin with four integers n,m,K,Q.
K is the number of Rook, Q is the number of queries.

Then K lines follow, each contain two integers x,y describing the coordinate of Rook.

Then Q lines follow, each contain four integers x1,y1,x2,y2 describing the left-down and right-up coordinates of query.

1≤n,m,K,Q≤100,000.

1≤x≤n,1≤y≤m.

1≤x1≤x2≤n,1≤y1≤y2≤m.

 

Output

For every query output "Yes" or "No" as mentioned above.

 

Sample Input

22 2 1 21 11 1 1 22 1 2 22 2 2 11 11 22 1 2 2

 

Sample Output

YesNoYes
Hint
Huge input, scanf recommended.

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>using namespace std;#define maxn 100007int tree[2][maxn];int main(){    int n,m,k,q,x,y,t,x2,y2;    scanf("%d",&t);    while(t--){        scanf("%d%d%d%d",&n,&m,&k,&q);        memset(tree,0,sizeof(tree));        for(int i = 0;i < k; i++){            scanf("%d%d",&x,&y);            tree[0][x] = 1;            tree[1][y] = 1;        }        for(int i = 1;i < maxn ;i++)            tree[0][i]+=tree[0][i-1],tree[1][i]+=tree[1][i-1];        while(q--){            scanf("%d%d%d%d",&x,&y,&x2,&y2);            if((x2-x+1==tree[0][x2]-tree[0][x-1]) || (y2-y+1==tree[1][y2]-tree[1][y-1]))                printf("Yes\n");            else                printf("No\n");        }    }    return 0;}