BZOJ3876 AHOI/JSOI2014支线剧情（上下界网络流）

　　原图所有边下界设为1上界设为inf花费为时间，那么显然就是一个上下界最小费用流了。做法与可行流类似。

　　因为每次选的都是最短路增广，且显然不会有负权增广路，所以所求出来的可行流的费用就是最小的。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<cstring>

#include<algorithm>

using namespace std;

int read()

{

    int x=,f=;char c=getchar();

    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}

    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();

    return x*f;

}

#define N 310

#define M 20010

#define inf 100000000

#define S 0

#define T 301

int n,p[N],t=-,pre[N],q[N],d[N],degree[N],ans=;

bool flag[N];

struct data{int to,nxt,cap,flow,cost;

}edge[M];

void addedge(int x,int y,int z,int cost)

{

    t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].cap=z,edge[t].flow=,edge[t].cost=cost,p[x]=t;

    t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=,edge[t].flow=,edge[t].cost=-cost,p[y]=t;

}

int inc(int &x){x++;if (x>n+) x-=n+;return x;}

bool spfa()

{

    memset(d,,sizeof(d));d[S]=;

    memset(flag,,sizeof(flag));

    int head=,tail=;q[]=S;

    do

    {

        int x=q[inc(head)];flag[x]=;

        for (int i=p[x];~i;i=edge[i].nxt)

        if (d[x]+edge[i].cost<d[edge[i].to]&&edge[i].flow<edge[i].cap)

        {

            d[edge[i].to]=d[x]+edge[i].cost;

            pre[edge[i].to]=i;

            if (!flag[edge[i].to])

            {

                q[inc(tail)]=edge[i].to;

                flag[edge[i].to]=;

            }

        }

    }while (head!=tail);

    return d[T]<inf;

}

void ekspfa()

{

    while (spfa())

    {

        int v=inf;

        for (int i=T;i!=S;i=edge[pre[i]^].to)

        v=min(v,edge[pre[i]].cap-edge[pre[i]].flow);

        for (int i=T;i!=S;i=edge[pre[i]^].to)

        ans+=v*edge[pre[i]].cost,edge[pre[i]].flow+=v,edge[pre[i]^].flow-=v;

    }

}

int main()

{

#ifndef ONLINE_JUDGE

    freopen("bzoj3876.in","r",stdin);

    freopen("bzoj3876.out","w",stdout);

    const char LL[]="%I64d";

#else

    const char LL[]="%lld";

#endif

    n=read();

    memset(p,,sizeof(p));

    for (int i=;i<=n;i++)

    {

        int k=read();

        if (i>) addedge(i,,inf,);

        while (k--)

        {

            int x=read(),y=read();

            addedge(i,x,inf,y);

            degree[i]++;degree[x]--;

            ans+=y;

        }

    }

    for (int i=;i<=n;i++)

    if (degree[i]>) addedge(i,T,degree[i],);

    else if (degree[i]<) addedge(S,i,-degree[i],);

    ekspfa();

    cout<<ans;

    return ;

}