BZOJ-3931 [CQOI2015]网络吞吐量 最短路 最大流

题面

题意:所有的流量都只能最短路径上的边,每个点有个最大的吞吐量,求1号点到n号点的最大吞吐量

题解:嗯嗯嗯 题意就是题解啊 

        先求最短路,然后可以是最短路上的边连容量inf的边

        每个点都拆点,然后除了1和n两个点,都,边就是他的吞吐量,1和n边当然是inf

        唯一注意就是long long

 

#include<bits/stdc++.h>
#define pa pair<int,int>
#define lld long long 
using namespace std;
#define N 1005
#define M 260000
#define inf 100000000000000LL
namespace Dinic
{
    int head[N],head2[N],p=1;
    struct Rec
    {
        int go,nex;
        lld c;
    }eg[M*2],e[M*2];
    void build(int a,int b,lld c)
    {
        eg[++p]=(Rec){b,head[a],-c};
        head[a]=p;
        eg[++p]=(Rec){a,head[b],0};
        head[b]=p;
    }
    lld dis[N],ans;
    int Q[N],s[N],S,T,stop;
    bool bfs()
    {
        memset(dis,0,sizeof(dis));
        dis[T]=1;
        Q[1]=T;
        for (int p1=1,p2=1;p1<=p2;p1++)
        {
            for (int i=head[Q[p1]];i;i=eg[i].nex)
                if (eg[i^1].c<0&&!dis[eg[i].go])
                {
                    dis[eg[i].go]=dis[Q[p1]]+1;
                    Q[++p2]=eg[i].go;
                }
        }
        if (!dis[S]) return false;
        memcpy(head2,head,sizeof(head));
        return true;
    }
    bool dinic(int p,int top)
    {
        if (p==T)
        {
            lld x=inf;
            for (int i=1;i<=top-1;i++) if (-eg[s[i]].c<x) x=-eg[s[i]].c,stop=i;
            for (int i=1;i<=top-1;i++) eg[s[i]].c+=x,eg[s[i]^1].c-=x;
            ans+=x;
            return true;
        }
        for (int &i=head2[p];i;i=eg[i].nex)
        {
            if (eg[i].c<0&&dis[eg[i].go]==dis[p]-1)
            {
                s[top]=i;
                if (dinic(eg[i].go,top+1)&&top!=stop) return true;
            }
        }
        return false;
    }
    lld ask()
    {
        ans=0;
        while (bfs()) dinic(S,1);
        return ans; 
    }
    void init(int _S,int _T){
        S=_S,T=_T;
    }
}
using namespace Dinic;
void clear()
{
    p=1;
    memset(head,0,sizeof(head));
}
int n,m,ss,tt,x[M],y[M];
long long xx,diss[N],z[M];
void addedge(int a,int b,lld c)
{
    p++;
    e[p].go=b;
    e[p].c=c;
    e[p].nex=head[a];
    head[a]=p;
}
void dijkstra()
{
    priority_queue<pa,vector<pa>,greater<pa> >q;
    for (int i=1;i<=n;i++) diss[i]=inf;
    diss[1]=0;
    q.push(make_pair(0,1));
    while (!q.empty())
    {
        int now=q.top().second;
        q.pop();
        for (int i=head[now];i;i=e[i].nex)
            if(diss[now]+e[i].c<diss[e[i].go])
            {
                diss[e[i].go]=diss[now]+e[i].c;
                q.push(make_pair(diss[e[i].go],e[i].go));
            }
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    ss=1;tt=n+n;
    init(ss,tt);
    for (int i=1;i<=m;i++)
    {
        scanf("%d%d%lld",&x[i],&y[i],&z[i]);
        addedge(x[i],y[i],z[i]);
        addedge(y[i],x[i],z[i]);
    }
    dijkstra();
    clear();
    for (int i=1;i<=m;i++)
    {
        if (diss[x[i]]+z[i]==diss[y[i]]) build(x[i]+n,y[i],inf);
        if (diss[y[i]]+z[i]==diss[x[i]]) build(y[i]+n,x[i],inf);
    }
    for (int i=1;i<=n;i++)
    {
        scanf("%lld",&xx);
        if (i==1 || i==n) build(i,i+n,inf);else build(i,i+n,xx);
    }
    printf("%lld",ask());
    return 0;
}