POJ 1149 PIGS Solution

比较垃圾所以刷的尽是些水题

Description
Mirko works on a pig farm that consists of M locked pig-houses and Mirko can’t unlock any pighouse because he doesn’t have the keys. Customers come to the farm one after another. Each of them has keys to some pig-houses and wants to buy a certain number of pigs.
All data concerning customers planning to visit the farm on that particular day are available to Mirko early in the morning so that he can make a sales-plan in order to maximize the number of pigs sold.
More precisely, the procedure is as following: the customer arrives, opens all pig-houses to which he has the key, Mirko sells a certain number of pigs from all the unlocked pig-houses to him, and, if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses.
An unlimited number of pigs can be placed in every pig-house.
Write a program that will find the maximum number of pigs that he can sell on that day.

Solution
首先我们想到把猪圈复制n遍然后建边。
最后在猪圈和人之间建边。
然而点太多Dinic跑不出来。
我们可以考虑缩掉一些点。
1）以后不向人连边的点就没必要复制了。
2）如果对于边c(u,v)=INF,那就可以把u,v给缩了
3）如果有些点他们的流量来源完全相同则可以缩掉。
4）如果有些点他们的流量流出完全相同则可以缩掉。
这样一通缩点以后我们可以惊奇地发现猪圈没了，只有人之间有边。
并且连边的规则是这样的：
对于一个人如果与他相连的猪圈没有被其他人连过则该人与src连一条容量为猪圈中猪的数量的边
不然就和上个连这个猪圈的人连一条容量为INF的边。
然后就可以过掉了。

Edelweiss神牛有一句很有意思的总结：
在面对网络流问题时，如果一时想不出很好的构图方法，不如先构造一个最直观，或者说最“硬来”的模型，然后再用合并节点和边的方式来简化这个模型。经过简化以后，好的构图思路就会涌现出来了。这是解决网络流问题的一个好方法。

[源代码]

#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>#include<queue>using namespace std;const int M=2e4+5;const int INF=1<<30;struct Edge{int to,cap,nxt;}edge[M<<1];int head[M],work[M],tot_edge,A[M],sink,src,last[M],dis[M],n,m;inline void add_edge(int from,int to,int c){    edge[tot_edge]=(Edge){to,c,head[from]};    head[from]=tot_edge++;    edge[tot_edge]=(Edge){from,0,head[to]};    head[to]=tot_edge++;}bool mark[M];inline int dfs(int v,int f){    if(v==sink)return f;    mark[v]=1;    for(int &i=work[v];~i;i=edge[i].nxt){        int to=edge[i].to;        if(!mark[to]&&dis[to]==dis[v]+1&&edge[i].cap){            int d=dfs(to,min(f,edge[i].cap));            if(d){                edge[i].cap-=d;                edge[i^1].cap+=d;                return d;            }        }    }    return 0;}queue<int>que;inline bool bfs(){    for(int i=1;i<=n;++i)        dis[i]=-1;    que.push(src);    dis[src]=0;    for(;!que.empty();){        int v=que.front();que.pop();        for(int i=head[v];~i;i=edge[i].nxt){            int to=edge[i].to;            if(dis[to]==-1&&edge[i].cap){                dis[to]=dis[v]+1;                que.push(to);            }        }    }    return dis[sink]!=-1;}inline int Dinic(){    int res=0;    for(;bfs();){        for(int i=1;i<=n;++i)            work[i]=head[i];        for(;;){            for(int i=1;i<=n;++i)mark[i]=0;            int f=dfs(src,INF);            if(!f)break;            res+=f;        }    }    return res;}int main(){    cin>>m>>n;    src=n+1,sink=n+2;    for(int i=1;i<=m;++i)        scanf("%d",&A[i]),last[i]=src;    memset(head,-1,sizeof(head));    for(int i=1,k,s;i<=n;++i){        scanf("%d",&k);        for(int j=1,a;j<=k;++j){            scanf("%d",&a);            if(last[a]!=src)add_edge(last[a],i,INF);            else add_edge(last[a],i,A[a]);            last[a]=i;        }        scanf("%d",&s);        add_edge(i,sink,s);    }    n+=2;    cout<<Dinic()<<endl;    return 0;}