Hdu 5480 Conturbatio
来源:互联网 发布:python数据分析 编辑:程序博客网 时间:2024/05/22 14:07
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?
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 integersn,m,K,Q .
K is the number of Rook, Q is the number of queries.
ThenK lines follow, each contain two integers x,y describing the coordinate of Rook.
ThenQ 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
Then
Then
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
YesNoYesHintHuge input, scanf recommended.
Source
BestCoder Round #57 (div.2)
Recommend
hujie | We have carefully selected several similar problems for you: 5493 5492 5491 5490 5489
在一个n×m的国际象棋棋盘上有很多车(Rook),其中车可以攻击他所属的一行或一列,包括它自己所在的位置。现在还有很多询问,每次询问给定一个棋盘内部的矩形,问矩形内部的所有格子是否都被车攻击到?
题解:记录下行攻击的前缀和及列攻击的前缀和,每个矩阵只要满足行攻击和列攻击全部攻击到就是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;}
0 0
- HDU 5480 Conturbatio
- Hdu 5480 Conturbatio
- hdu 5480Conturbatio
- HDU 5480 Conturbatio
- hdu 5480 Conturbatio(水)
- hdu 5480 Conturbatio【模拟】
- hdu 5480 Conturbatio
- 【HDU 5480 Conturbatio】
- HDU 5480 Conturbatio
- HDU Problem 5480 Conturbatio
- hdu 5480 Conturbatio 线段树
- HDU 5480:Conturbatio 前缀和
- 【前缀和】hdu 5480 Conturbatio
- hdu 5480 Conturbatio 区间和
- HDU 5480 Conturbatio 树状数组
- HDU 5480 Conturbatio (前缀和)
- 前缀和 hdu 5480 (Conturbatio)
- HDU 5480 Conturbatio(前缀和)
- hdu acm 1880 魔咒词典
- Yes,搞定---关于CreateRemoteThread引起对象进程崩溃的处理
- linux内核同步方法
- WebService初步了解
- test 我的文章为什么没有了
- Hdu 5480 Conturbatio
- string stringbuffer stringbuilder区别
- 为什么我delete后内存没有被释放?
- 【javaScript】基础知识
- 聊天室功能实现代码
- nyoj808蚂蚁的难题(八)【dp】
- Socket--java网络编程
- BFC环境(block formatting context块级格式化上下文)
- 简易计算器