1 #pragma comment(linker,"/STACK:100000000,100000000")//阔栈的语句 2 #include <stdio.h> 3 #include <string.h> 4 #include <vector> 5 #define pb push_back 6 using namespace std; 7 const int maxn = 200002; 8 const int maxm = 1000002; 9 struct EDGE{10     int next, to, vis,w;11 }edge[maxm*2];//记录边，vis表示这一条边是否被访问12 struct GEEDGE {13     int u, to;14 }geedge[maxm];//记录割边15 int n,m,time,top,w,type,getot,cnt;//time是时间戳,type标记连通块,getot割边的数量，cnt边数16 int head[maxn],st[maxn],dfn[maxn],low[maxn],belo[maxn];//belo[i],表示i属于哪个连通块17 int start,ans1,ans,u,v,tval[maxn];18 void init(){19     time=top=type=getot=cnt=0;20     memset(head,-1,sizeof(head));21     memset(dfn,0,sizeof(dfn));22     memset(low,0,sizeof(low));23     memset(belo,0,sizeof(belo));24 }25 void add(int u,int v,int w){26    edge[cnt].to=v;27    edge[cnt].next=head[u];28    edge[cnt].vis=0;29    edge[cnt].w=w;30    head[u]=cnt++;31 }32 void dfs(int u) {//可以处理重边的情况33     low[u] = dfn[u] = ++time;34     st[++top] = u;35     for(int i = head[u];i != -1;i = edge[i].next) {36         if(edge[i].vis)    continue;37         edge[i].vis = edge[i^1].vis = 1;38         int to = edge[i].to;39         if(!dfn[to]) {40             dfs(to);41             low[u] = min(low[u], low[to]);42             if(low[to] > dfn[u]) {//表示找到一个连通块43                 type++;44                 int v;45                 do {46                     v = st[top--];47                     belo[v] = type;//表示v属于第type个连通块48                 } while(v != to);49                 geedge[++getot].u = u;//记录割边的起点50                 geedge[getot].to = to;//记录割边的另一个点51                 tval[getot]=edge[i].w;52             }53         }54         else55             low[u] = min(low[u], low[to]);56     }57 }58 int main(){59      while(~scanf("%d%d",&n,&m)&&n){60          init();61          for(int i=1;i<=m;i++){62             scanf("%d%d %d",&u,&v,&w);63             add(u,v,w);64             add(v,u,w);65          }66          dfs(1);bool flag=false;67          for(int i=1;i<=n;i++){68             if(!dfn[i]){69                 flag=true;70                 break;71             }72          }73          ans1=getot;ans=100000000;74             for(int i=1;i<=ans1;i++){75                 ans=min(ans,tval[i]);76             }77          if(ans==0)ans=1;78          if(flag)ans=0;79          if(!ans1&&!flag)printf("-1\n");80          else81         printf("%d\n",ans);82      }83 }