HDU4738-Caocao's Bridges

Caocao's Bridges

                                                                              Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                        Total Submission(s): 4186    Accepted Submission(s): 1315

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 <= N ² )

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
  

 

Source

2013 ACM/ICPC Asia Regional Hangzhou Online

 

Recommend

liuyiding

 

题意：有n个点，m个桥，求出所有的桥，然后输出桥的权值的最小值，没有桥输出-1。
解题思路：如果一开始是不连通的，输出0；图有重边，需要处理，如果取到的最小值是0的话，要输出1，表示要派一个人过去。

#include <iostream>  
#include <cstdio>  
#include <string>  
#include <cstring>  
#include <algorithm>  
#include <cmath>  
#include <queue>  
#include <vector>  
#include <set>  
#include <stack>  
#include <map>  
#include <climits>  
  
using namespace std;  
  
#define LL long long  
const int INF=0x3f3f3f3f;  
const int N=1010;  
  
struct Edge  
{  
    int v,w,flag,nt;  
} edge[N*1000];  
int s[N],cnt,ans,sum;  
int n,m;  
int dfn[N],low[N],dep;  
bool vis[N];  
  
void AddEdge(int u,int v,int w)  
{  
    for(int i=s[u];~i;i=edge[i].nt)  
    {  
        if(edge[i].v==v)  
        {  
            edge[i].flag++;  
            return ;  
        }  
    }  
    edge[cnt].v=v;  
    edge[cnt].nt=s[u];  
    edge[cnt].flag=0;  
    edge[cnt].w=w;  
    s[u]=cnt++;  
}  
  
void tarjan(int u,int pre)  
{  
    vis[u]=true;  
    dfn[u]=low[u]=++dep;  
    for(int i=s[u];~i;i=edge[i].nt)  
    {  
        int v=edge[i].v;  
        if(v==pre) continue;  
        if(!vis[v])  
        {  
            tarjan(v,u);  
            low[u]=min(low[u],low[v]);  
        }  
        else low[u]=min(low[u],dfn[v]);  
        if((low[v]>dfn[u])&&!edge[i].flag)  
            ans=min(ans,edge[i].w);  
    }  
}  
  
int main()    
{    
    while(~scanf("%d%d",&n,&m)&&(n+m))    
    {    
        memset(s,-1,sizeof s);    
        cnt=sum=0;  
        ans=INF;  
        for(int i=0; i<m; i++)    
        {    
            int u,v,w;    
            scanf("%d%d%d",&u,&v,&w);    
            AddEdge(u,v,w);
			AddEdge(v,u,w);
        }    
        memset(vis,0,sizeof vis);
        dep=0; 
        for(int i=1; i<=n; i++)    
            if(!vis[i]) tarjan(i,0),sum++;    
        // Debug    
        /* for(int i = 1; i <= n; i++)  
             printf("dfn[%d] = %d, low[%d] = %d\n", i,dfn[i], i,low[i]);  
         for(int i = 1; i <= n; i++)  
             printf("id[%d] = %d\n", i, id[i] );*/    
	     if(ans==0) ans=1;  
	     if(ans==INF) ans=-1;  
	     if(sum>1) ans=0;  
	     printf("%d\n",ans);  
    }    
    return 0;    
}