POJ1149 PIGS 【最大流 + 构图】

题目链接：http://poj.org/problem?id=1149

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions:24094		Accepted: 10982

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.

题目大意：

1.给定m个猪棚，n个顾客，接下来n行每行代表第i个顾客有k个猪棚的钥匙，以及该顾客的需求量。

2.注意题目的要求，主人可以在顾客打开了猪棚的时候对猪棚中的猪数量进行调动，所以，为了尽可能满足所有顾客的需求，建图需要体现对猪的调动。

3.对于每一个猪棚，第一个打开它的顾客，我们将源点向该顾客建边，容量为猪棚的猪数量。对于以后再次光顾该猪棚的顾客，我们将第一次光顾该猪棚的顾客与以后光顾该猪棚的顾客建边，容量为inf。最后将每个顾客与终点连边，边的容量为顾客的需求量。至于为什么这样建图，可以理解为将猪全部给了第一个光顾该猪棚的顾客，以后再需要的顾客可以从他这里拿，这样的话可以确保尽可能多的满足后来的顾客。

代码如下：

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 #include<queue>
 #define mem(a, b) memset(a, b, sizeof(a))
 using namespace std;
 const int MAXM = ; //猪圈数目上界
 const int MAXN = ; //顾客数目上界
 const int inf = 0x3f3f3f3f;

 int m, n; //猪棚数目 顾客数目
 int num[MAXM], first[MAXM], vis[MAXM];//每个猪圈里猪的数目 每个猪圈第一个顾客 每个猪圈是否已经有第一个顾客买了
 int head[MAXN], cnt;
 int dep[MAXN];
 queue<int> Q;

 struct Edge
 {
     int to, next, flow;
 }edge[ * MAXN +  * MAXN * MAXN];

 void add(int a, int b, int c)
 {
     cnt ++;
     edge[cnt].to = b;
     edge[cnt].flow = c;
     edge[cnt].next = head[a];
     head[a] = cnt;
 }

 int bfs(int st, int ed)
 {
     if(st == ed)
         return ;
     while(!Q.empty())    Q.pop();
     mem(dep, -);
     dep[st] = ;
     Q.push(st);
     while(!Q.empty())
     {
         int index = Q.front();
         Q.pop();
         for(int i = head[index]; i != -; i = edge[i].next)
         {
             int to = edge[i].to;
             if(edge[i].flow >  && dep[to] == -)
             {
                 dep[to] = dep[index] + ;
                 Q.push(to);
             }
         }
     }
     return dep[ed] != -;
 }

 int dfs(int now, int ed, int zx)
 {
     if(now == ed)
         return zx;
     for(int i = head[now]; i != -; i = edge[i].next)
     {
         int to = edge[i].to;
         if(dep[to] == dep[now] +  && edge[i].flow > )
         {
             int flow = dfs(to, ed, min(zx, edge[i].flow));
             if(flow > )
             {
                 edge[i].flow -= flow;
                 edge[i ^ ].flow += flow;
                 return flow;
             }
         }
     }
     return -;
 }

 void dinic(int st, int ed)
 {
     int ans = ;
     while(bfs(st, ed))
     {
         while()
         {
             int inc = dfs(st, ed, inf);
             if(inc == -)
                 break;
             ans += inc;
         }
     }
     printf("%d\n", ans);
 }

 int main()
 {
     int m, n;
     scanf("%d%d", &m, &n);
     int st = , ed = n + ;
     mem(head, -), cnt = -;
     mem(first, -), mem(vis, );
     for(int i = ; i <= m; i ++)
         scanf("%d", &num[i]);
     for(int i = ; i <= n; i ++)
     {
         int k;
         scanf("%d", &k);
         for(int j = ; j <= k; j ++)
         {
             int x;
             scanf("%d", &x);//猪圈编号
             if(!vis[x])
             {
                 first[x] = i;
                 vis[x] = ;
                 add(st, i, num[x]);
                 add(i, st, );
             }
             else
             {
                 add(first[x], i, inf);
                 add(i, first[x], );
             }
         }
         int x;
         scanf("%d", &x);
         add(i, ed, x);
         add(ed, i, );
     }
     dinic(st, ed);
     return ;
 }

POJ1149