1122 Hamiltonian Cycle （25 分）
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 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, 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题意：
给出一个图，并给出每次询问，判断每次询问的路径是否为哈密尔顿回路
思路：
先建图，然后用bool的vis数组判断是否访问过了，终点另外判断（稍微模拟的有一些麻烦）
#include<iostream>
#include<set>
#include<cstring>
using namespace std;
const int maxn=250;
const int INF =1000000;
int g[maxn][maxn]={INF};
int road[maxn];
bool vis[maxn]={false};
set<int>s;
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++)
{
int a,b;
cin>>a>>b;
g[a][b]=1;
g[b][a]=1;
}
int k;
scanf("%d",&k);
while(k--)
{
int num;
memset(vis,false,sizeof(vis));
s.clear();
cin>>num;
int st;
scanf("%d",&st);
int temp=st;//记录起点
vis[st]=true;
int flag=0;
s.insert(st);
for(int i=1;i<num-1;i++)
{
int x;
scanf("%d",&x);
//起点到x有路且x没有访问过
if(g[st][x]!=1||vis[x]==true)
{
flag=1;
continue;
}
s.insert(x);
st=x;
}
//cout<<s.size()<<endl;
int end;
scanf("%d",&end);
if(g[st][end]!=1||end!=temp)
flag=1;
if(s.size()!=n)
flag=1;
if(flag==1)
printf("NO\n");
else
printf("YES\n");
}
}