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=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),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=-1,pre[N],q[N],d[N],degree[N],ans=0;
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=0,edge[t].cost=cost,p[x]=t;
    t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=0,edge[t].flow=0,edge[t].cost=-cost,p[y]=t;
}
int inc(int &x){x++;if (x>n+1) x-=n+1;return x;}
bool spfa()
{
    memset(d,42,sizeof(d));d[S]=0;
    memset(flag,0,sizeof(flag));
    int head=0,tail=1;q[1]=S;
    do
    {
        int x=q[inc(head)];flag[x]=0;
        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]=1;
            }
        }
    }while (head!=tail);
    return d[T]<inf;
}
void ekspfa()
{
    while (spfa())
    {
        int v=inf;
        for (int i=T;i!=S;i=edge[pre[i]^1].to)
        v=min(v,edge[pre[i]].cap-edge[pre[i]].flow);
        for (int i=T;i!=S;i=edge[pre[i]^1].to)
        ans+=v*edge[pre[i]].cost,edge[pre[i]].flow+=v,edge[pre[i]^1].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,255,sizeof(p));
    for (int i=1;i<=n;i++)
    {
        int k=read();
        if (i>1) addedge(i,1,inf,0);
        while (k--)
        {
            int x=read(),y=read();
            addedge(i,x,inf,y);
            degree[i]++;degree[x]--;
            ans+=y;
        }
    }
    for (int i=1;i<=n;i++)
    if (degree[i]>0) addedge(i,T,degree[i],0);
    else if (degree[i]<0) addedge(S,i,-degree[i],0);
    ekspfa();
    cout<<ans;
    return 0;
}