poj 1149 PIGS（最大流）

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PIGS

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 16815 Accepted: 7551

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.

Input

The first line of input contains two integers M and N, 1 <= M <= 1000, 1 <= N <= 100, number of pighouses and number of customers. Pig houses are numbered from 1 to M and customers are numbered from 1 to N. 
The next line contains M integeres, for each pig-house initial number of pigs. The number of pigs in each pig-house is greater or equal to 0 and less or equal to 1000. 
The next N lines contains records about the customers in the following form ( record about the i-th customer is written in the (i+2)-th line): 
A K1 K2 ... KA B It means that this customer has key to the pig-houses marked with the numbers K1, K2, ..., KA (sorted nondecreasingly ) and that he wants to buy B pigs. Numbers A and B can be equal to 0.

Output

The first and only line of the output should contain the number of sold pigs.

Sample Input

3 33 1 102 1 2 22 1 3 31 2 6

Sample Output

7

题意就是卖猪  有m个猪圈  每个猪圈里面有一些猪  

有顾客过来买猪  顾客拥有猪圈的钥匙  可以打开猪圈  然后买该猪圈里面的猪

猪可以在从一个猪圈移到另外一个猪圈  

卖猪的想要卖出最多的猪  求做多猪的数量

建图：

设立一个源点和一个汇点

将源点与每个猪圈的第一个顾客之间建边  权值为该猪圈中猪的数量 如果出现重边  权值相加

将每个顾客与汇点之间建边 权值为顾客买猪的数量

对于同一猪圈 上一个顾客与下一个顾客之间建边 权值为INF

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <string.h>#include <string>#include <vector>#include <queue>#define eps 1e-8#define op operator#define MOD  10009#define MAXN  1100#define INF 30000000#define MEM(a,x)    memset(a,x,sizeof a)#define ll __int64using namespace std;struct Dinic{    struct Edge    {        int from,to,cap,flow;        Edge(){}        Edge(int from,int to,int cap,int flow):from(from),to(to),cap(cap),flow(flow){}    };    vector<Edge> edges;    vector<int> G[MAXN];    bool vis[MAXN];    int d[MAXN];    int cur[MAXN];    int n,m,s,t,maxflow;    void init(int n)    {        this->n=n;        for(int i=0;i<=n;i++)            G[i].clear();        edges.clear();    }    void addedge(int from,int to,int cap)    {        edges.push_back(Edge(from,to,cap,0));        edges.push_back(Edge(to,from,0,0));        m=edges.size();        G[from].push_back(m-2);        G[to].push_back(m-1);    }    bool bfs()    {        MEM(vis,0);        MEM(d,-1);        queue<int> q;        q.push(s);        d[s]=maxflow=0;        vis[s]=1;        while(!q.empty())        {            int u=q.front(); q.pop();            int sz=G[u].size();            for(int i=0;i<sz;i++)            {                Edge e=edges[G[u][i]];                if(!vis[e.to]&&e.cap>e.flow)                {                    d[e.to]=d[u]+1;                    vis[e.to]=1;                    q.push(e.to);                }            }        }        return vis[t];    }    int dfs(int u,int a)    {        if(u==t||a==0)  return a;        int sz=G[u].size();        int flow=0,f;        for(int &i=cur[u];i<sz;i++)        {            Edge &e=edges[G[u][i]];            if(d[u]+1==d[e.to]&&(f=dfs(e.to,min(a,e.cap-e.flow)))>0)            {                e.flow+=f;                edges[G[u][i]^1].flow-=f;                flow+=f;                a-=f;                if(a==0)  break;            }        }        return flow;    }    int Maxflow(int s,int t)    {        this->s=s;  this->t=t;        int flow=0;        while(bfs())        {            MEM(cur,0);            flow+=dfs(s,INF);        }        return flow;    }}Dic;int main(){//freopen("ceshi.txt","r",stdin);    int n,m;    while(scanf("%d%d",&m,&n)!=EOF)    {        int c[MAXN];//各个猪圈的容量        int last[MAXN];//各个猪圈上一次打开的人        MEM(last,0);        for(int i=1;i<=m;i++)            scanf("%d",&c[i]);        Dic.init(n+1);        for(int i=1;i<=n;i++)        {            int num;            scanf("%d",&num);            for(int j=0;j<num;j++)            {                int x;                scanf("%d",&x);                if(last[x]==0)                {                    last[x]=i;                    Dic.addedge(0,i,c[x]);                }                else                {//                    last[x]=i;                    Dic.addedge(last[x],i,INF);                    last[x]=i;                }            }            int y;            scanf("%d",&y);            Dic.addedge(i,n+1,y);        }        printf("%d\n",Dic.Maxflow(0,n+1));    }    return 0;}

//以上的代码很好的解决重边问题  出现重边 直接建边