PAT A1122 hamiltonian cycle (25)

//哈密顿环路，能够沿边形成一个覆盖所有点的环路，中间点不重复。判断：1、k=n+1 2、数组首尾相同才能形成环路 3、每个点出现1次 4、必须有边才能走

#include<cstdio>#include<vector>#include<algorithm>using namespace std;const int MAX=210;int G[MAX][MAX]={0};int judge(vector<int> vn){int vis[MAX]={0};int i;if(vn[0]!=vn[vn.size()-1]) return 0;//not a cyclefor(i=1;i<vn.size();i++){if(G[vn[i-1]][vn[i]]==1){//if has an edgevis[vn[i]]++;if(vis[vn[i]]>1) return 0;}else return 0;//not a edge}return 1;}int main(){int n,m,i;scanf("%d%d",&n,&m);int v1,v2;for(i=0;i<m;i++){scanf("%d%d",&v1,&v2);G[v1][v2]=G[v2][v1]=1;}int k,temp,query;scanf("%d",&query);for(i=0;i<query;i++){scanf("%d",&k);vector<int> vn;for(int j=0;j<k;j++){scanf("%d",&temp);vn.push_back(temp);}if(k==(n+1)){if(judge(vn)==1) printf("YES\n");else printf("NO\n");}else printf("NO\n");}system("pause");return 0;}