Hdu 5480 Conturbatio

Conturbatio

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 341    Accepted Submission(s): 155

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

2 2 2 1 2 1 1 1 1 1 2 2 1 2 2 2 2 2 1 1 1 1 2 2 1 2 2

 

Sample Output

Yes No Yes

Hint

Huge input, scanf recommended.

 

Source

BestCoder Round #57 (div.2)

 

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在一個$n\times m$的國際象棋棋盤上有很多車(Rook)，其中車可以攻擊他所屬的一行或一列，包括它自己所在的位置。
現在還有很多詢問，每次詢問給定一個棋盤內部的矩形，問矩形內部的所有格子是否都被車攻擊到？

題解：記錄下行攻擊的前綴和及列攻擊的前綴和，每個矩陣只要滿足行攻擊和列攻擊全部攻擊到就是yes.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <set>
#include <queue>

using namespace std;
typedef long long ll;
const int maxn=1e5+10;

int n,m,k,q;
int cs[maxn],rs[maxn];

void init() {
    memset(cs,0,sizeof cs);//hang
    memset(rs,0,sizeof rs);//lie
}
int main() {
    int t;
    cin>>t;
    int ca=1;
    while(t--) {
        scanf("%d%d%d%d",&n,&m,&k,&q);
        init();
        for(int i=0; i<k; i++) {
            int x,y;
            scanf("%d%d",&x,&y);
            cs[x]=1;
            rs[y]=1;
        }
        for(int i=1; i<=n; i++)
            cs[i]+=cs[i-1];
        for(int i=1; i<=m; i++)
            rs[i]+=rs[i-1];
        while(q--) {
            int lx,ly,rx,ry;
            scanf("%d%d%d%d",&lx,&ly,&rx,&ry);
            if(cs[rx]-cs[lx-1]==rx-lx+1||(rs[ry]-rs[ly-1]==ry-ly+1)) {
                printf("Yes\n");
            } else {
                printf("No\n");
            }
        }
    }
    return 0;
}