hdu_5889_Barricade(最小割+最短路)

题目链接：hdu_5889_Barricade

题意：

有n个点，m条边，每个边的长度都为1，每个边有一个消耗w，如果要阻断这条路，那么就会消耗w，现在让你阻断点1到点n的所有最短路，问你最小的消耗是多少

题解：

先用dij算出最短路，然后再枚举每一条边，如果dis[u]+1=dis[v]，那么久在网络流里加一条u到v的边，消耗为w，

最后用板子跑一下最大流就行了

#include<bits/stdc++.h>
#define F(i,a,b) for(int i=a;i<=b;++i)
using namespace std;

#define MAXN 20010//边数
#define inf 10000000
int t,n,m,x,y,c,egg[MAXN][3];
int v[MAXN], w[MAXN], nxt[MAXN], gg[MAXN], ed, dd[MAXN];//n为点数,d为起点到每点的最短路程，初始化ed为0,g初始化为0
void adg(int x, int y, int z) { v[++ed] = y; w[ed] = z; nxt[ed] = gg[x]; gg[x] = ed;}
typedef pair<int, int>P;
priority_queue<P, vector<P>, greater<P> > Q;
void dijkstra(int S) {
    int i, x;
    for (i = 1; i <= n; i++)dd[i] = inf; Q.push(P(dd[S] = 0, S));
    while (!Q.empty()) {
        P t = Q.top(); Q.pop();
        if (dd[x = t.second] < t.first)continue;
        for (i = gg[x]; i; i = nxt[i])if (dd[x] + w[i] < dd[v[i]])Q.push(P(dd[v[i]] = dd[x] + w[i], v[i]));
    }
}

const int N=2000,M=20010;
struct edge{int t,f;edge*nxt,*pair;}*g[N],*d[N],pool[M],*cur=pool;
struct ISAP{
    int n,m,i,S,T,h[N],gap[N],maxflow;
    void init(int ss,int tt){for(S=ss,T=tt,cur=pool,i=1;i<=T;i++)g[i]=d[i]=NULL,h[i]=gap[i]=0;}
    void add(int s,int t,int f){
        edge*p=cur++;p->t=t,p->f=f,p->nxt=g[s],g[s]=p;
        p=cur++,p->t=s,p->f=0,p->nxt=g[t],g[t]=p;
        g[s]->pair=g[t],g[t]->pair=g[s];
    }
    int sap(int v,int flow){
        if(v==T)return flow;
        int rec=0;
        for(edge*p=d[v];p;p=p->nxt)if(h[v]==h[p->t]+1&&p->f){
        int ret=sap(p->t,min(flow-rec,p->f));
        p->f-=ret;p->pair->f+=ret;d[v]=p;
        if((rec+=ret)==flow)return flow;
        }
        if(!(--gap[h[v]]))h[S]=T;
        gap[++h[v]]++;d[v]=g[v];
        return rec;
    }
    int get_ans(){
        for(gap[maxflow=0]=T,i=1;i<=T;i++)d[i]=g[i];
        while(h[S]<T)maxflow+=sap(S,inf);
        return maxflow;
    }
}G;

int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        G.init(1,n);
        memset(gg,0,sizeof(gg)),ed=0;
        F(i,1,m)
        {
            scanf("%d%d%d",&x,&y,&c),adg(x,y,1),adg(y,x,1);
            egg[i][0]=x,egg[i][1]=y,egg[i][2]=c;
        }
        dijkstra(1);
        F(i,1,m)
        {
            if(dd[egg[i][0]]+1==dd[egg[i][1]])G.add(egg[i][0],egg[i][1],egg[i][2]);
            if(dd[egg[i][1]]+1==dd[egg[i][0]])G.add(egg[i][1],egg[i][0],egg[i][2]);
        }
        printf("%d
",G.get_ans());
    }
}