Conturbatio
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 156 Accepted Submission(s): 79
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)
/*
可以发现如果一个矩阵被全部攻击到,
很显然要么是因为它的每一行都有车,
或者每一列都有车。
所以只需要记录一下哪些行和哪些列有车,
对于每个询问只需要做一个前缀和就可以知道答案了。
*/
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define inf -0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
int x[101000],y[100010];
int main(){
int t;
scanf("%d",&t);
int n,m,K,Q;
int x1,x2,y1,y2;
while(t--){
scanf("%d%d%d%d",&n,&m,&K,&Q);
mem0(x);
mem0(y);
while(K--){
scanf("%d%d",&x1,&y1);
x[x1]=1;
y[y1]=1;
}
for(int i=1;i<=n;i++)
x[i]+=x[i-1];
for(int j=1;j<=m;j++)
y[j]+=y[j-1];
while(Q--){
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
if(x[x2]-x[x1-1]==x2-x1+1||y[y2]-y[y1-1]==y2-y1+1)
printf("Yes\n");
else
printf("No\n");
}
}
return 0;
}