HDU Problem 5480 Conturbatio

Conturbatio

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 981    Accepted Submission(s): 442

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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很水的一道题。如果棋盘上的一个点会受到攻击，那么他所在的列和行上最少有一个车，也就是说一个车就确定了一行一列。用两个数组存放从第一行（列）到当前行（列）有多少行（列）会受到攻击就可以了。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <vector>
#include <cstdlib>
using namespace std;
//typedef long long LL;
//typedef __int64 Int;
typedef pair<int,int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double Pi = acos(-1);
const int MOD = 1e9+5;
const int MAXN = 100000 + 10;
int visx[MAXN], visy[MAXN];
int main() {
    int T, m, n, K, Q, a1, b1, a2, b2, a, b;
    scanf("%d", &T);
    while (T--) {
        memset(visx, 0, sizeof(visx));
        memset(visy, 0, sizeof(visy));
        scanf("%d%d%d%d", &n, &m, &K, &Q);
        for (int i = 0; i < K; i++) {
            scanf("%d%d", &a, &b);
            visx[a] = visy[b] = 1;
        }
        for (int i = 1; i <= n; i++) {
            visx[i] += visx[i - 1];
        }
        for (int i = 1; i <= m; i++) {
            visy[i] += visy[i - 1];
        }
        while (Q--) {
            scanf("%d%d%d%d", &a1, &b1, &a2, &b2);
            if (visx[a2] - visx[a1 - 1] == (a2 - a1 + 1))
                printf("Yes
");
            else if (visy[b2] - visy[b1 - 1] == (b2 - b1 + 1))
                printf("Yes
");
            else printf("No
");
        }
    }
    return 0;
}