hdoj5480 Conturbatio(思维)

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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
有一个n*m的棋盘，有k个车（和象棋里的车一样），有q次询问，问车是否能将给出的矩形的所有格子打到，给出车的坐标，矩形左下角的坐标和右上角的坐标.cow[i]表示以第i行结尾的矩形，col[i]表示以第i列结尾的矩形.
代码如下：
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int cow[100010];int col[100010];int main(){int n,m,k,q;int t;scanf("%d",&t);while(t--){scanf("%d%d%d%d",&n,&m,&k,&q);memset(cow,0,sizeof(cow));memset(col,0,sizeof(col));int i;int x,y;for(i=0;i<k;i++){scanf("%d%d",&x,&y);cow[x]=col[y]=1;}for(i=1;i<=n;i++){if(cow[i])cow[i]+=cow[i-1];}for(i=1;i<=m;i++){if(col[i])col[i]+=col[i-1];}int x1,x2,y1,y2;for(i=1;i<=q;i++){scanf("%d%d%d%d",&x1,&y1,&x2,&y2);if(cow[x2]>=x2-x1+1||col[y2]>=y2-y1+1)printf("Yes\n");elseprintf("No\n");}}}

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