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

题目大意：

给定n个点和m条边  和每条边的价值，求桥的最小价值（最小桥）

看着挺简单的但是有好多细节：

1、会有重边

2、如果最小价值是0的话应该输出1

3、m条边有可能不能连通n个点，这个时候没有花费。

Caocao‘s Bridges

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3108    Accepted Submission(s): 982

Problem Description

Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But he wouldn‘t give up. Caocao‘s army still was not good at water battles, so he came up with another idea. He built many islands in the Changjiang river, and based on those islands, Caocao‘s army could easily attack Zhou Yu‘s troop. Caocao also built bridges connecting islands. If all islands were connected by bridges, Caocao‘s army could be deployed very conveniently among those islands. Zhou Yu couldn‘t stand with that, so he wanted to destroy some Caocao‘s bridges so one or more islands would be seperated from other islands. But Zhou Yu had only one bomb which was left by Zhuge Liang, so he could only destroy one bridge. Zhou Yu must send someone carrying the bomb to destroy the bridge. There might be guards on bridges. The soldier number of the bombing team couldn‘t be less than the guard number of a bridge, or the mission would fail. Please figure out as least how many soldiers Zhou Yu have to sent to complete the island seperating mission.

Input

There are no more than 12 test cases.

In each test case:

The first line contains two integers, N and M, meaning that there are N islands and M bridges. All the islands are numbered from 1 to N. ( 2 <= N <= 1000, 0 < M <= N² )

Next M lines describes M bridges. Each line contains three integers U,V and W, meaning that there is a bridge connecting island U and island V, and there are W guards on that bridge. ( U ≠ V and 0 <= W <= 10,000 )

The input ends with N = 0 and M = 0.

Output

For each test case, print the minimum soldier number Zhou Yu had to send to complete the mission. If Zhou Yu couldn‘t succeed any way, print -1 instead.

Sample Input

3 3
1 2 7
2 3 4
3 1 4
3 2
1 2 7
2 3 4
0 0

Sample Output

-1
4

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>

using namespace std;
#define N 1010
#define INF 0x3f3f3f3f

struct node
{
    int to,flew,next;
}edge[N*N];

int low[N],dfn[N],Time,top,ans,Stack[N],belong[N],sum,head[N];

void Inn()
{
    memset(low,0,sizeof(low));
    memset(dfn,0,sizeof(dfn));
    memset(Stack,0,sizeof(Stack));
    memset(belong,0,sizeof(belong));
    memset(head,-1,sizeof(head));
    Time=top=ans=sum=0;
}

void add(int from,int to,int flew)
{
    edge[ans].to=to;
    edge[ans].flew=flew;
    edge[ans].next=head[from];
    head[from]=ans++;
}
void Tarjin(int u,int f)
{
    int v,k=0;
    low[u]=dfn[u]=++Time;
    Stack[top++]=u;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        v=edge[i].to;
        if(v==f && !k)
        {
            k++;
            continue;
        }
        if(!dfn[v])
        {
            Tarjin(v,u);
            low[u]=min(low[u],low[v]);
        }
        else
            low[u]=min(low[u],dfn[v]);
    }
    if(dfn[u]==low[u])
    {
        sum++;
        do
        {
            v=Stack[--top];
            belong[v]=sum;
        }while(v!=u);
    }

}

void solve(int n)
{
    int k=0;
    int Min=INF;
    for(int i=1;i<=n;i++)
    {
        if(!dfn[i])
        {
            k++;
            Tarjin(i,0);
        }
    }
    if(sum==1)
    {
        printf("-1\n");
        return;
    }
    if(k>1)
    {
        printf("0\n");
        return;
    }
        for(int i=1;i<=n;i++)
        {
            for(int j=head[i];j!=-1;j=edge[j].next)
            {
                int u=belong[i];
                int v=belong[edge[j].to];
                if(u!=v)
                {
                    Min=min(Min,edge[j].flew);
                }
            }
        }
    if(Min==0)
        Min++;
        printf("%d\n",Min);
}
int main()
{
    int n,m,a,b,c,i;
    while(scanf("%d %d",&n,&m),n+m)
    {
        Inn();
        for(i=0;i<m;i++)
        {
            scanf("%d %d %d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
        }
        solve(n);
    }
    return 0;
}

Caocao's Bridges-HDU4738(Tarjin+求桥)