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 = 1100; //猪圈数目上界 
const int MAXN = 110; //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[4 * MAXN + 2 * 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 0;
    while(!Q.empty())    Q.pop();
    mem(dep, -1);
    dep[st] = 1;
    Q.push(st);
    while(!Q.empty())
    {
        int index = Q.front();
        Q.pop();
        for(int i = head[index]; i != -1; i = edge[i].next)
        {
            int to = edge[i].to;
            if(edge[i].flow > 0 && dep[to] == -1)
            {
                dep[to] = dep[index] + 1;
                Q.push(to);
            }
        }
    }
    return dep[ed] != -1;
}

int dfs(int now, int ed, int zx)
{
    if(now == ed)
        return zx;
    for(int i = head[now]; i != -1; i = edge[i].next)
    {
        int to = edge[i].to;
        if(dep[to] == dep[now] + 1 && edge[i].flow > 0)
        {
            int flow = dfs(to, ed, min(zx, edge[i].flow));
            if(flow > 0)
            {
                edge[i].flow -= flow;
                edge[i ^ 1].flow += flow;
                return flow;
            }
        }
    }
    return -1;
}

void dinic(int st, int ed)
{
    int ans = 0;
    while(bfs(st, ed))
    {
        while(1)
        {
            int inc = dfs(st, ed, inf);
            if(inc == -1)
                break;
            ans += inc;
        }
    }
    printf("%d
", ans);
}

int main()
{
    int m, n;
    scanf("%d%d", &m, &n);
    int st = 0, ed = n + 1;
    mem(head, -1), cnt = -1;
    mem(first, -1), mem(vis, 0);
    for(int i = 1; i <= m; i ++)
        scanf("%d", &num[i]);
    for(int i = 1; i <= n; i ++)
    {
        int k;
        scanf("%d", &k);
        for(int j = 1; j <= k; j ++)
        {
            int x;
            scanf("%d", &x);//猪圈编号 
            if(!vis[x])
            {
                first[x] = i;
                vis[x] = 1;
                add(st, i, num[x]);
                add(i, st, 0);
            }
            else
            {
                add(first[x], i, inf);
                add(i, first[x], 0);
            }
        }
        int x;
        scanf("%d", &x);
        add(i, ed, x);
        add(ed, i, 0);
    }
    dinic(st, ed);
    return 0;
}