情况分好即可 可以设置三个布尔变量分别代表是否为环，是否为简单环 两个节点之间是否有边；

#include <bits/stdc++.h>
using namespace std;
int graph[205][205];
int N,M,K,T;
int main(){
    cin>>N>>M;
    while (M--){
        int a,b,c;
        cin>>a>>b>>c;
        graph[a][b]=graph[b][a]=c;
    }
    cin>>K;
    int shortIndex=-1,shortDis=INT_MAX;
    for(int k=1;k<=K;k++){
        cin>>T;
        int A[T],dis=0;
        bool cycle=true,Na= false,simple= true;
        unordered_map<int,int> m;
        for (int i = 0; i < T; ++i) {
            cin>>A[i];
            if (i>0){
                m[A[i]]++;
                Na=graph[A[i-1]][A[i]]==0? true:Na;
                dis+=graph[A[i-1]][A[i]];
            }
        }
        cycle=!Na and A[T-1]==A[0] and m.size()==N?cycle:false;
        for(auto &it:m) {
            if (it.second > 1)
                simple = false;
        }
        printf("Path %d: ",k);
        if (Na)
            printf("NA ");
        else {
            printf("%d ", dis);
        }
        printf("(%s)\n",cycle?(simple?"TS simple cycle":"TS cycle"):"Not a TS cycle");

        if (cycle and shortDis>dis){
            shortDis=dis;
            shortIndex=k;
        }
    }
    printf("Shortest Dist(%d) = %d",shortIndex,shortDis);
    return 0;
}