思路：
首先 先Dijkstra一遍 找出来最短路
不是最短路上的边都不要

然后呢 套个Dinic模板就好了……

求个最小割

输出

大功告成~~

//By SiriusRen
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 22222
int n,m,xx,yy,zz,first[N],tot,dis[N],vis[N],ans,cases;
struct Edge{int v,w,next;}edge[N];
struct Node{int now,weight;}jy;
void add(int xx,int yy,int zz){
    edge[tot].v=yy,edge[tot].w=zz;
    edge[tot].next=first[xx],first[xx]=tot++;
}
bool operator < (Node a,Node b){return a.weight>b.weight;}
void Dijkstra(int from){
    memset(vis,0,sizeof(vis));
    memset(dis,0x3f,sizeof(dis));
    dis[from]=0;
    priority_queue<Node>pq;
    jy.now=from,jy.weight=0;
    pq.push(jy);
    while(!pq.empty()){
        Node t=pq.top();pq.pop();
        if(vis[t.now])continue;
        vis[t.now]=1;
        for(int i=first[t.now];~i;i=edge[i].next)
            if(!vis[edge[i].v]&&dis[edge[i].v]>dis[t.now]+1){
                dis[edge[i].v]=dis[t.now]+1;
                jy.now=edge[i].v;jy.weight=t.weight+1;
                pq.push(jy);
            }
    }
}
struct Dinic{
    int v[N],next[N],w[N],fst[N],cnt;
    void init(){memset(fst,-1,sizeof(fst)),cnt=0;}
    void add(int x,int y,int z){
        w[cnt]=z,v[cnt]=y;
        next[cnt]=fst[x],fst[x]=cnt++;
    }
    bool tell(){
        memset(vis,-1,sizeof(vis));
        queue<int>q;
        q.push(1),vis[1]=0;
        while(!q.empty()){
            int t=q.front();q.pop();
            for(int i=fst[t];~i;i=next[i]){
                if(w[i]&&vis[v[i]]==-1)
                    q.push(v[i]),vis[v[i]]=vis[t]+1;
            }
        }
        return vis[n]!=-1;
    }
    int zeng(int x,int y){
        if(x==n||!y)return y;
        int r=0;
        for(int i=fst[x];~i&&y>r;i=next[i])
            if(w[i]&&vis[v[i]]==vis[x]+1){
                int t=zeng(v[i],min(y-r,w[i]));
                w[i]-=t,w[i^1]+=t,r+=t;
            }
        if(!r)vis[x]=-1;
        return r;
    }
    void flow(){
        while(tell())while(xx=zeng(1,0x3fffffff))ans+=xx;
        printf("%d\n",ans);
    }
}dinic;
int main(){
    scanf("%d",&cases);
    while(cases--){
        memset(first,-1,sizeof(first)),tot=ans=0;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&xx,&yy,&zz);
            add(xx,yy,zz),add(yy,xx,zz);
        }
        Dijkstra(1),dinic.init();
        for(int i=1;i<=n;i++)
            for(int j=first[i];~j;j=edge[j].next)
                if(dis[edge[j].v]==dis[i]+1){
                    dinic.add(i,edge[j].v,edge[j].w);
                    dinic.add(edge[j].v,i,0);
                }
        dinic.flow();
    }
}

HDU 5889 Barricade （Dijkstra+Dinic）_最短路