Barricade

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Description

The empire is under attack again. The general of empire is planning to defend his castle. The land can be seen as N towns and M roads, and each road has the same length and connects two towns. The town numbered 1 is where general‘s castle is located, and the town numbered N is where the enemies are staying. The general supposes that the enemies would choose a shortest path. He knows his army is not ready to fight and he needs more time. Consequently he decides to put some barricades on some roads to slow down his enemies. Now, he asks you to find a way to set these barricades to make sure the enemies would meet at least one of them. Moreover, the barricade on the i-th road requires wi units of wood. Because of lacking resources, you need to use as less wood as possible.

Input

The first line of input contains an integer t, then t test cases follow.
For each test case, in the first line there are two integers N(N≤1000) and M(M≤10000).
The i-the line of the next M lines describes the i-th edge with three integers u,v and w where 0≤w≤1000 denoting an edge between u and v of barricade cost w.

Output

For each test cases, output the minimum wood cost.

Sample Input

1

4 4

1 2 1

2 4 2

3 1 3

4 3 4

Sample Output

4

/*
 * HDU 5889 Barricade
 * 求堵住从1到n的最短路的最小花费
 *
 * bfs + 最小割
 * 首先用Dij或者bfs求出dis数组，接着从终点出发，倒着bfs找出所有最短路径重新建图
 * 然后就是一个裸的最小割了，就等于最大流
 */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
#include <map>
#include <vector>
using namespace std;

struct NetFlow
{
    const static int MAXN = 5000+10;
    const static int MAXE = 500000;
    const static long long INF = 1e18;
    struct Edge
    {
        int from,to,next,cap,flow;
        Edge(){}
        Edge(int u,int v,int c,int f,int nxt):from(u),to(v),cap(c),flow(f),next(nxt) {}
    }edge[MAXE];
    int head[MAXN],tol,N;
    int cur[MAXN],pre[MAXN],dis[MAXN],gap[MAXN];
    void init(int _n)
    {
        N=_n,tol=0; memset(head,-1,sizeof(head));
    }
    void link(int u,int v,int cap)//s->t,cap
    {
        edge[tol]=Edge(u,v,cap,0,head[u]);head[u]=tol++;
        edge[tol]=Edge(v,u,0,0,head[v]);head[v]=tol++;
    }
    long long ISAP(int S,int T)
    {//S -> T
        long long maxflow=0,aug=INF;
        int flag=false,u,v;
        for (int i=0;i<N;++i) cur[i]=head[i],gap[i]=dis[i]=0;
        for (gap[S]=N,u=pre[S]=S;dis[S]<N;flag=false)
        {
            for (int &it=cur[u];it!=-1;it=edge[it].next)
            {
                if (edge[it].cap>edge[it].flow&&dis[u]==dis[v=edge[it].to]+1)
                {
                    aug=min(aug,(long long)(edge[it].cap-edge[it].flow));
                    pre[v]=u,u=v; flag=true;
                    if (u==T)
                    {
                        for(maxflow+=aug;u!=S;)
                        {
                            edge[cur[u=pre[u]]].flow+=aug;
                            edge[cur[u]^1].flow-=aug;
                        }
                        aug=INF;
                    }
                    break;
                }
            }
            if(flag) continue;
            int mx=N;
            for(int it=head[u];it!=-1;it=edge[it].next)
            {
                if(edge[it].cap>edge[it].flow&&dis[edge[it].to]<mx)
                {
                    mx=dis[edge[it].to]; cur[u]=it;
                }
            }
            if((--gap[dis[u]])==0) break;
            ++gap[dis[u]=mx+1]; u=pre[u];
        }
        return maxflow;
    }
}NF;
const int MAXN = 5000+10;
const int MAXE = 200000;
const int INF = 1e9;
struct node{
    int v,c;
    node(int _v=0,int _c=0):v(_v),c(_c){}
    bool operator <(const node &rhs) const{
        return c>rhs.c;
    }
};
struct Ed
{
    int u,v,w;
}e;
vector<Ed>p;
struct Edge{
    int to,cost;
    int w;
    int next;
};
Edge edge[MAXE];
int head[MAXN],tot;
bool vis[MAXN];
int dis[MAXN];

void Dijkstra(int n,int start)
{
    memset(vis,false,sizeof(vis));
    for(int i=1;i<=n;i++) dis[i]=INF;
    priority_queue<node>q;
    while(!q.empty()) q.pop();
    dis[start]=0;
    q.push(node(start,0));
    node next;
    while(!q.empty()){
        next=q.top();
        q.pop();
        int u=next.v;
        if(vis[u]) continue;
        vis[u]=true;
        for(int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].to;
            int cost=edge[i].cost;
            if(!vis[v]&&dis[v]>dis[u]+cost){
                dis[v]=dis[u]+cost;
                q.push(node(v,dis[v]));
            }
        }
    }
}

void init()
{
    tot=0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int w,int ww)
{
    edge[tot].to=v;
    edge[tot].cost=w;
    edge[tot].w=ww;
    edge[tot].next=head[u];
    head[u]=tot++;
}
void bfs(int T)
{
    memset(vis,0,sizeof(vis));
    queue<int>Q;
    Q.push(T);
    vis[T]=true;
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int ds=edge[i].cost;
            int v=edge[i].to;
            if(ds+dis[v]==dis[u])
            {
                e.u=v,e.v=u,e.w=edge[i].w;
                p.push_back(e);
                if(!vis[v])
                {
                    vis[v]=1;
                    Q.push(v);
                }
            }
        }
    }
}

int main()
{
    int T;
    int n,m,u,v,w;
    //freopen("in.txt","r",stdin);
    scanf("%d",&T);
    while(T--)
    {
        init();
        p.clear();
        scanf("%d%d",&n,&m);
        while(m--)
        {
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,1,w);
            addedge(v,u,1,w);
        }
        Dijkstra(n,1);
        bfs(n);
        int sz=p.size();
        NF.init(n+1);
        for(int i=0;i<sz;i++)
        {
            u=p[i].u;
            v=p[i].v;
            w=p[i].w;
            NF.link(u,v,w);
        }
        long long ans=NF.ISAP(1,n);
        printf("%lld\n",ans);
    }
    return 0;
}