解析：

在一個無向圖裡找出一個由至少三個點組成環，使得環上邊的權值和最小。

程式碼：

#include<cctype>
#include<cstdio>
#include<algorithm>
#include<cstring> 
using namespace std;
#define ll long long
#define re register
#define gc getchar
#define pc putchar
#define cs const
#define rank Rank

inline
ll getint(){
    re ll num=0
 
;
    re char c;
    while(!isdigit(c=gc()));
    while(isdigit(c))num=(num<<1)+(num<<3)+(c^48),c=gc();
    return num;
}

inline
void outint(ll a){
    static char ch[23];
    if(a==0)pc('0');
    while(a)ch[++ch[0]]=(a-a/10*10)^48,a/=10;
    while(ch[0])pc(ch[ch[0]--]);
}

int n,m;
int ans=0x3f3f3f3f
 
;
int dist[101][101];
int a[101][101];

signed main(){
    while(~scanf("%d%d",&n,&m)){
        memset(a,0x3f,sizeof a);
        ans=0x3f3f3f3f;
        for(int re i=1;i<=m;++i){
            int u=getint(),v=getint(),val=getint();
            a[u][v]=a[v][u]=min(a[u][v],val);
        }

        memcpy
 
(dist,a,sizeof a);

        for(int re k=1;k<=n;++k){
            for(int re i=1;i<k;++i){
                for(int re j=i+1;j<k;++j){
                    if(ans>(ll)dist[i][j]+a[j][k]+a[k][i]){
                        ans=dist[i][j]+a[j][k]+a[k][i];
                    }
                }
            }
            for(int re i=1;i<=n;++i){
                if(i!=k)
                for(int re j=1;j<=n;++j){
                    if(j!=k&&j!=i)
                    if(dist[i][j]>dist[i][k]+dist[k][j]){
                        dist[i][j]=dist[i][k]+dist[k][j];
                     }
                }
            }
        }
        if(ans==0x3f3f3f3f)puts("It's impossible.");
        else outint(ans),pc('\n');
    }
    return 0;
}