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

題意：給出 n 個點 m 條邊的無向圖，m 行資料中，前兩個代表相連的點，後一個代表這條邊的價值，現要取掉一條邊使圖變成不連通的，求要花費的最小价值，如果無法使圖變為不連通圖，則輸出 -1

思路：實質是 Tarjan 演算法求割邊，但要輸出邊價值最小的割邊，用 vector 做比較麻煩，因此考慮使用鄰接表來做

要注意的是，如果無向圖本身就已經是不連通的了，直接輸出 0 即可，由於有重邊的情況，因此要在 Tarjan 演算法時處理重邊，由於存在權值為 0 的情況，因此如果求出來的最小權值橋的權值為 0 時，也要排一個人去炸橋，此時輸出 1

Source Program

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 20001
#define MOD 16007
#define E 1e-6
#define LL long long
using namespace std;
vector<int> G[N];
int n,m;
int dfn[N],low[N];
int head[N];
int edge_cnt;
int block_cnt;
int res;
struct Edge{
    int to;//下一個點
    int next;//下一跳邊
    int id;//增邊的順序編號
    int cost;//邊的權值
}edge[N];
void add(int x,int y,int cost,int id){
    edge[edge_cnt].to=y;
    edge[edge_cnt].cost=cost;
    edge[edge_cnt].id=id;
    edge[edge_cnt].next=head[x];
    head[x]=edge_cnt++;
}
void Tarjan(int x,int father){
    low[x]=dfn[x]=++block_cnt;

    for(int i=head[x];~i;i=edge[i].next){
        int y=edge[i].to;
        int id=edge[i].id;

        if(id==father)
            continue;

        if(dfn[y]==0){
            Tarjan(y,id);
            low[x]=min(low[x],low[y]);

            if(dfn[x]<low[y])
                res=min(res,edge[i].cost);
        }
        else
            low[x]=min(low[x],dfn[y]);
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF&&(n+m)){
        edge_cnt=0;
        memset(head,-1,sizeof(head));
        for(int i=1;i<=m;i++){
            int x,y,w;
            scanf("%d%d%d",&x,&y,&w);
            add(x,y,w,i);
            add(y,x,w,i);

        }

        res=INF;
        block_cnt=0;
        memset(dfn,0,sizeof(dfn));

        int connection=0;
        for(int i=1;i<=n;i++){
            if(dfn[i]==0){
                connection++;
                Tarjan(i,0);
            }
        }

        if(res==0)
            res=1;
        if(res==INF)
            res=-1;
        if(connection>1)
            res=0;

        printf("%d\n",res);
    }
    return 0;
}