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=1010;
const int INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-6;
const double OO=1e60;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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[2*N*N];
bool iscut[2*N*N];
int first[N],next[2*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=0;
    for(i=first[u];i!=-1;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=-1;
        bcnt++;
        while(x!=u){
            x=s.top();s.pop();
            bccno[x]=bcnt;
        }
    }
}

int find_bcc()
{
    int i,cnt=0;
    bcnt=dfs_clock=0;
    mem(pre,0);mem(bccno,0);mem(iscut,0);
    for(i=1;i<=n;i++){
        if(!pre[i]){cnt++;dfs(i,-1);}
    }
    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,-1);mt=0;
        for(i=0;i<m;i++){
            scanf("%d%d%d",&a,&b,&c);
            adde(a,b,c);
        }

        t=find_bcc();
        if(t>1){
            printf("0
");
            continue;
        }

        ans=INF;
        for(i=0;i<mt;i++){
            if(iscut[i])ans=Min(ans,e[i].w);
        }
        printf("%d
",ans==INF?-1:(ans?ans:1));
    }
    return 0;
}