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?
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.
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
Recommend
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。
copy
#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\n"); else if (visy[b2] - visy[b1 - 1] == (b2 - b1 + 1)) printf("Yes\n"); else printf("No\n"); } } return 0; }
本文作者: zprhhs
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