hdu 4738 Caocao's Bridges -有重邊的無向圖找橋

Caocao's Bridges

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5576    Accepted Submission(s): 1751

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

/**
無向圖，有重邊的情況找最小的橋
題解：要消除反向邊對其的影響，但是不能消除重邊對其的影響，
因爲存圖時是按照正向邊反向邊存的，所以第一次遇到v==fa的一定是因爲反向邊而存下來的
*/
#include<iostream>
#include<cstdio>
#include<vector>
#include<sstream>
#include<cstring>
#include<algorithm>
#define mem(a,x) memset(a,x,sizeof(a))
#define INF 0x3f3f3f3f
using namespace std;
const int N = 1010;
struct node{
    int v,w,id;
    node(int v = 0,int w = 0,int id = 0):v(v),w(w),id(id){};
};
vector<node>G[N];
int pre[N];
int low[N];
int dfsLock;int ans ;int n,m;
void init(){
    mem(pre,0); mem(low,0);
    for(int i=0;i<=n;i++) G[i].clear();
    dfsLock = 0;ans = INF;
}
int dfs(int u,int fa){
    low[u] = pre[u] = ++dfsLock;
    int k = 1;
    for(int i=0;i<G[u].size();i++){
        int v = G[u][i].v;

        if(v == fa && k) {///沒有重邊則不需要這句話
            k = 0;continue;
        }

        if(!pre[v]){
            dfs(v,u);
            low[u] = min(low[u],low[v]);///用後代的low更新當前的
            if(low[v] > pre[u])
                ans = min(ans,G[u][i].w);
        }
        else
            low[u] = min(low[u],pre[v]);///利用後代v的反向邊更新low
    }
}
int main(){
    int t;
    while(scanf("%d%d",&n,&m)!=EOF&& (n || m)){
        int a,b,c;
        init();
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&a,&b,&c);
            G[a].push_back(node(b,c,i));
            G[b].push_back(node(a,c,i));
        }
        int cnt = 0;
        for(int i=1;i<=n;i++){
            if(!pre[i]){
                cnt++;///連通圖的個數
                dfs(i,-1);///根沒有父親，傳-1
            }
        }
        if(cnt > 1) ///多於一個聯通分量 不用派人
            ans = 0;
        else if(ans == INF)
            ans = -1;
        else if(ans == 0) ///沒有人看守，也需要派一個人去炸橋
            ans = 1;
        printf("%d\n",ans);
    }
    return 0;
}