Time Limit: 40000/20000 MS (Java/Others)
Memory Limit: 32768/132768 K (Java/Others)

Problem Description
Degree, King led Meow bear ha Group warriors, ready to attack the group crashed.
crashed the family is a powerful nation, there are full of wisdom counselors have infinite power of the soldiers.
so that a war will be very difficult.
for better offensive crashed family, we should first decide the degree of collapse crashed bear from inside the group.
should be such that the first step is not concentric inner crashed Group Qi, the need for internal gap.
a total aromatic crashed $n$nTwo generals, a total of them $m$mA strong relationship, destroy every strong relationship requires a certain price.
Degree, now you need to destroy some of the commands bears a strong relationship, such that the interior of the generals, these can not be a strong relationship, even as a full communication block, to ensure smooth war.
should at least ask how much to pay.

Input
The title comprising a plurality of sets of test data.
first line two integers $n，m$n，m, Expressed $n$nA general. $m$mA relationship.
Next $m$mRows of three integers $u,v,w$u,v,w. Indicates the presence of a strong relationship between u and v generals generals, to destroy the strong relationship comes at a cost $w$w

data range:
$2<=n<=3000$2<=n<=3000
$1<=m<=100000$1<=m<=100000
$1<=u,v<=n$1<=u,v<=n
$1<=w<=1000$1<=w<=1000

Output
For each test, the cost required minimum output.

Sample Input

2 1
1 2 1
3 3
1 2 5
1 2 4
2 3 3

Sample Output

1
3

answer:
global minimum cut bare title

Global minimum cut of implementation:
1. selecting a smallest first pair of points in the s-t G is cut, update your answers.
2. The selected s-t merge into a collection.
3. do repeated until only one vertex G.

How to find the minimum in the current state of the cut?
1. (after collection or combined) to pick a starting point.
2. In this starting point, through all points to this (set) by the other adjacent dots (set)
3. root with a large collection heap memory is connected directly to the last added last remaining point and the point that the final expansion union to a set inside the line.

#include<bits/stdc++.h>
#define ll long long
#define pa pair<ll,int>
using namespace std;
const ll INF=1e18;
struct edge{
    int to,w;
};
int n,m;
vector<edge>G[3004];
int f[3004],lk[3004];
int Find(int x){return f[x]==x?x:f[x]=Find(f[x]);}
void Union(int x,int y){
    int p=x;
    while(lk[p])p=lk[p];
    lk[p]=y;
    f[y]=x;
}
ll val[3004];
bool vis[3004];
priority_queue<pa>q;
ll msp(int cnt,int &s,int &t){
    for(int i=1;i<=n;i++){
        vis[i]=0;
        val[i]=0;
    }
    while(!q.empty())q.pop();
    t=1;
    while(--cnt){
        vis[t]=1;
        s=t;
        for(int x=s;x;x=lk[x]){
            for(int i=0;i<G[x].size();i++){
                int to=Find(G[x][i].to),w=G[x][i].w;
                if(!vis[to]){
                    val[to]+=w;
                    q.push({val[to],to});
                }
            }
        }
        t=0;
        while(!t){
            if(q.empty())return 0;
            ll ft=q.top().first;
            int st=q.top().second;
            q.pop();
            if(val[st]==ft){
                t=st;
            }
        }
    }
    return val[t];
}
ll StoerWanger(){
    ll ans=INF;
    int s,t;
    for(int i=n;i>1;i--){
        ans=min(ans,msp(i,s,t));
        if(ans==0)break;
        Union(s,t);
    }
    return ans;
}
int w33ha(){
    for(int i=1;i<=n;i++){
        f[i]=i;
        lk[i]=0;
        G[i].clear();
    }
    for(int i=1;i<=m;i++){
        int u,v,w;scanf("%d%d%d",&u,&v,&w);
        G[u].push_back((edge){v,w});
        G[v].push_back((edge){u,w});
    }
    printf("%lld\n",StoerWanger());
    return 0;
}
int main(){
    while(scanf("%d%d",&n,&m)!=EOF)w33ha();
    return 0;
}