PAT_A1122#Hamiltonian Cycle
Source:
Description:
The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a graph. Such a cycle is called a "Hamiltonian cycle".
In this problem, you are supposed to tell if a given cycle is a Hamiltonian cycle.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive integers N (2), the number of vertices, and M, the number of edges in an undirected graph. Then Mlines follow, each describes an edge in the format
Vertex1 Vertex2
, where the vertices are numbered from 1 to N. The next line gives a positive integer K which is the number of queries, followed by Klines of queries, each in the format:n V1 V2 ... Vn
where n is the number of vertices in the list, and Vi's are the vertices on a path.
Output Specification:
For each query, print in a line
YES
if the path does form a Hamiltonian cycle, orNO
if not.
Sample Input:
6 10
6 2
3 4
1 5
2 5
3 1
4 1
1 6
6 3
1 2
4 5
6
7 5 1 4 3 6 2 5
6 5 1 4 3 6 2
9 6 2 1 6 3 4 5 2 6
4 1 2 5 1
7 6 1 3 4 5 2 6
7 6 1 2 5 4 3 1
Sample Output:
YES
NO
NO
NO
YES
NO
Keys:
Code:
/*
Data: 2019-06-20 16:30:23
Problem: PAT_A1122#Hamiltonian Cycle
AC: 12:29 题目大意:
H圈定义:简单圈且包含全部顶点;
判断所给圈是否为H圈
*/
#include<cstdio>
#include<set>
#include<algorithm>
using namespace std;
const int M=;
int grap[M][M],path[M]; int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("Test.txt", "r", stdin);
#endif // ONLINE_JUDGE fill(grap[],grap[]+M*M,);
int n,m,k,v1,v2;
scanf("%d%d", &n,&m);
for(int i=; i<m; i++)
{
scanf("%d%d", &v1,&v2);
grap[v1][v2]=;
grap[v2][v1]=;
}
scanf("%d", &m);
while(m--)
{
set<int> ver;
scanf("%d", &k);
for(int i=; i<k; i++){
scanf("%d", &path[i]);
ver.insert(path[i]);
}
int reach=;
for(int i=; i<k-; i++){
if(grap[path[i]][path[i+]]==){
reach=;break;
}
}
if(reach== || path[]!=path[k-] || ver.size()!=n || k!=n+)
printf("NO\n");
else
printf("YES\n");
} return ;
}
PAT_A1122#Hamiltonian Cycle的更多相关文章
- PAT1122: Hamiltonian Cycle
1122. Hamiltonian Cycle (25) 时间限制 300 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue The ...
- A1122. Hamiltonian Cycle
The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a gra ...
- PAT A1122 Hamiltonian Cycle (25 分)——图遍历
The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a gra ...
- 1122 Hamiltonian Cycle (25 分)
1122 Hamiltonian Cycle (25 分) The "Hamilton cycle problem" is to find a simple cycle that ...
- PAT甲级 1122. Hamiltonian Cycle (25)
1122. Hamiltonian Cycle (25) 时间限制 300 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue The ...
- hihoCoder-1087 Hamiltonian Cycle (记忆化搜索)
描述 Given a directed graph containing n vertice (numbered from 1 to n) and m edges. Can you tell us h ...
- PAT 1122 Hamiltonian Cycle[比较一般]
1122 Hamiltonian Cycle (25 分) The "Hamilton cycle problem" is to find a simple cycle that ...
- PAT 1122 Hamiltonian Cycle
The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a gra ...
- 1122. Hamiltonian Cycle (25)
The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a gra ...
随机推荐
- android NDK开发在本地C/C++源码中设置断点单步调试具体教程
近期在学android NDK开发,折腾了一天,最终可以成功在ADT中设置断点单步调试本地C/C++源码了.网上关于这方面的资料太少了,并且大都不全,并且调试过程中会出现各种各样的问题,真是非常磨人. ...
- oracle中使用impdp数据泵导入数据提示“ORA-31684:对象类型已经存在”错误的解决
转载请注明出处:http://blog.csdn.net/dongdong9223/article/details/47448751 本文出自[我是干勾鱼的博客] oracle中使用impdp数据泵导 ...
- servlet中的中文乱码问题
老师总会说道:学完这个知识点,我们来谈谈中文乱码问题. 乱码的问题总是无处不在,处理不好会给用户带极差的用户体验. 那么我们来记录一下servlet中的乱码问题吧! 1.服务器向客户端响应时出现的乱码 ...
- Privoxy shadowscocks代理
ubuntu已经启动好了sock5的代理, 代理为: 127.0.0.1:1080. #使用Privoxy将sock5代理映射为http代理. 安装Privoxy sudo apt-get updat ...
- 电脑升级win10后visio的问题
上周由于电脑意外蓝屏,系统从win7升级到了win10,昨天工作写文档时才发现缺少画图的工具,于是按照了visio2013,在编辑设计图时发现,一旦用visio打开或编辑图后再到word里设计图的内容 ...
- 使用filezella服务器配置ftp
使用FileZilla配置FTP站点,可参考以下步骤: 1.打开Filezilla Server服务端: 点击[Edit]->[Users],或者点击如下图标新增用户. 2.添加FTP帐号后,设 ...
- golang单点推送
package main import ( "encoding/json" "flag" "fmt" "log" &qu ...
- POJ 3635 优先队列BFS
(感谢lyd学长的幻灯片) 注意vis数组的应用 在vis[i][j]中 i表示到了第i个点 j表示还剩j升油 vis[i][j]表示最小话费. 这样只需搜到话费比它少的更新入堆就OK了 //By: ...
- java.util.Date
package com.etc.usual; import java.util.Calendar; import java.util.Date; /** * * @author Administrat ...
- String和八种基本数据类型互相转换
//String转换为对应的八种基本数据类型 String str="100"; //Value out of range. Value:"200" Radix ...