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
- Spring MVC的@RequestMapping多个URL映射到同一个方法
@RequestMapping可以是一个URL对应一个方法,也可以多个URL对应同一个方法,写法如下: @RequestMapping(value={"url","res ...
- N天学习一个linux命令之ip
用途 show / manipulate routing, devices, policy routing and tunnels 用法 通用格式 ip [ OPTIONS ] OBJECT { CO ...
- 以"小刀会“的成败论当今创业成败
讲起"小刀会",熟悉的人或许非常熟悉,不熟悉的人或许根本不知道清末有这样一个组织. 依据翻查史料,最初的小刀会是在福建成立的,来源有两个.一个是天地会的分支,一个是白莲教分支. 而 ...
- UVa 642 - Word Amalgamation
题目:给你一个单词列表.再给你一些新的单词.输出列表中又一次排列能得到此新单词的词. 分析:字符串.对每一个字符串的字母排序生成新的传f(str).总体排序,用二分来查找就可以. 说明:注意输出要满足 ...
- poj 1664 放苹果 (划分数)
题意:中文题目,不解释... 题解: 第一种方法是暴力深搜:枚举盘子1~n放苹果数量的所有情况,不需要剪枝:将每次枚举的情况,即每个盘的苹果数量,以字典序排序,然后存进set里 以此去重像" ...
- Test for Job (poj 3249 记忆化搜索)
Language: Default Test for Job Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 9733 A ...
- 使用Memcached改进Java企业级应用性能:架构和设置
Memcached由Danga Interactive开发.用来提升LiveJournal.com站点性能. Memcached分布式架构支持众多的社交网络应用,Twitter.Facebook还有W ...
- hdu 6082 度度熊与邪恶大魔王(2017"百度之星"程序设计大赛 - 资格赛 )
度度熊与邪恶大魔王 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total S ...
- POJ - 3281 Dining(拆点+最大网络流)
Dining Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 18230 Accepted: 8132 Descripti ...
- 【POJ 3630】 Phone List
[题目链接] http://poj.org/problem?id=3630 [算法] 字典树 [代码] #include <algorithm> #include <bitset&g ...