HDU-4738 Caocao's Bridges 边联通分量

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4738

　　题意：在有重边的无向图中，求权值最小的桥。

　　注意trick就好了，ans为0时输出1，总要有一个人去丢炸弹吧。。。

 //STATUS:C++_AC_62MS_8144KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 #pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e60;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 /* Edge-Biconnected Component(可以处理重边)
   iscut[]为割边集
   bccno[]为双连通点集,保存为编号        */
 struct Edge{
     int u,v,w;
 }e[*N*N];
 bool iscut[*N*N];
 int first[N],next[*N*N],pre[N],low[N],bccno[N];
 int n,m,mt,bcnt,dfs_clock;
 stack<int> s;

 void adde(int a,int b,int c)
 {
     e[mt].u=a;e[mt].v=b;e[mt].w=c;
     next[mt]=first[a];first[a]=mt++;
     e[mt].u=b;e[mt].v=a;e[mt].w=c;
     next[mt]=first[b];first[b]=mt++;
 }

 void dfs(int u,int fa)
 {
     int i,v;
     pre[u]=low[u]=++dfs_clock;
     s.push(u);
     int cnt=;
     for(i=first[u];i!=-;i=next[i]){
         v=e[i].v;
         if(!pre[v]){
             dfs(v,u);
             low[u]=Min(low[u],low[v]);
             if(low[v]>pre[u])iscut[i]=true;   //存在割边
         }
         else if(fa==v){  //反向边更新
             if(cnt)low[u]=Min(low[u],pre[v]);
             cnt++;
         }
         else low[u]=Min(low[u],pre[v]);
     }
     if(low[u]==pre[u]){  //充分必要条件
         int x=-;
         bcnt++;
         while(x!=u){
             x=s.top();s.pop();
             bccno[x]=bcnt;
         }
     }
 }

 int find_bcc()
 {
     int i,cnt=;
     bcnt=dfs_clock=;
     mem(pre,);mem(bccno,);mem(iscut,);
     for(i=;i<=n;i++){
         if(!pre[i]){cnt++;dfs(i,-);}
     }
     return cnt;
 }

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,a,b,c,ans,t;
     while(~scanf("%d%d",&n,&m) && (n||m))
     {
         mem(first,-);mt=;
         for(i=;i<m;i++){
             scanf("%d%d%d",&a,&b,&c);
             adde(a,b,c);
         }

         t=find_bcc();
         if(t>){
             printf("0\n");
             continue;
         }

         ans=INF;
         for(i=;i<mt;i++){
             if(iscut[i])ans=Min(ans,e[i].w);
         }
         printf("%d\n",ans==INF?-:(ans?ans:));
     }
     return ;
 }