A1122 Hamiltonian Cycle (25 point(s))

简单环判断

1. 原文

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<N≤200), the number of vertices, and M, the number of edges in an undirected graph. Then M lines 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 K lines 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, or NO 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

2. 解析

路径判断

if(graph[s][t]==Inf)

简单判断

if(visit[t]==true)

环判断

if(visit[i]==false)
if(start!=t)

3. AC代码

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=210;
const int Inf=312312312;
int graph[maxn][maxn];
bool visit[maxn]={false};
int main()
{
	fill(graph[0],graph[0]+maxn*maxn,Inf);
	int n,m;
	scanf("%d%d",&n,&m);
	int a,b;
	for (int i = 0; i < m; ++i)
	{
		scanf("%d%d",&a,&b);
		graph[a][b]=graph[b][a]=1;
	}
	int k;
	scanf("%d",&k);
	while(k--)
	{
		fill(visit,visit+maxn,false);
		bool simple=true; bool cycle=true; bool connected=true;
		int start,s,t; int a;
		scanf("%d%d",&a,&s);
		start=s;
		for (int i = 1; i < a; ++i)
		{
			scanf("%d",&t);
			if (visit[t]==true)
			{
				simple=false;
			}
			visit[t]=true;
			if (graph[s][t]==Inf)
			{
				connected=false;
			}
			s=t;
		}
		for (int i = 1; i <= n; ++i)
		{
			if (visit[i]==false)
			{
				cycle=false;
			}
		}
		//printf("%d %d %d %d\n", start!=t,!simple,!cycle,!connected);
		if (start!=t||!simple||!cycle||!connected)
		{
			printf("NO\n");
		}
		else{
			printf("YES\n");
		}
	}

	return 0;
}