POJ 1149 PIGS（最大流）

题目链接：http://poj.org/problem?id=1149

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

Source

Croatia OI 2002 Final Exam - First day

【题目大意】 （转：http://wenku.baidu.com/view/0ad00abec77da26925c5b01c.html）
有M个猪圈，每个猪圈里初始时有若干头猪。一开始所有猪圈都是关闭的。依次来了N个顾客，每个顾客分别会打开指定的几个猪圈，从中买若干头猪。每个顾客分别都有他能够买的数量的上限。每个顾客走后，他打开的那些猪圈中的猪，都可以被任意地调换到其它开着的猪圈里，然后所有猪圈重新关上。问总共最多能卖出多少头猪。（1 <= N <= 100, 1 <= M <= 1000）  
举个例子来说。有3个猪圈，初始时分别有3、1和10头猪。依次来了3个顾客，第一个打开1号和2号猪圈，最多买2头；第二个打开1号和3号猪圈，最多买3头；第三个打开2号猪圈，最多买6头。那么，最好的可能性之一就是第一个顾客从1号圈买2头，然后把1号圈剩下的1头放到2号圈；第二个顾客从3号圈买3头；第三个顾客从2号圈买2头。总共卖出2+3+2=7头。 
 
【建模方法】 
不难想象，这个问题的网络模型可以很直观地构造出来。就拿上面的例子来说，可以构造出图1所示的模型（图中凡是没有标数字的边，容量都是∞）： 
• 三个顾客，就有三轮交易，每一轮分别都有3个猪圈和1个顾客的结点。 • 从源点到第一轮的各个猪圈各有一条边，容量就是各个猪圈里的猪的初始数量。 
• 从各个顾客到汇点各有一条边，容量就是各个顾客能买的数量上限。 • 在某一轮中，从该顾客打开的所有猪圈都有一条边连向该顾客，容量都是∞。 
• 最后一轮除外，从每一轮的i号猪圈都有一条边连向下一轮的i号猪圈，容量都是∞，表示这一轮剩下的猪可以留到下一轮。 
• 最后一轮除外，从每一轮被打开的所有猪圈，到下一轮的同样这些猪圈，两两之间都要连一条边，表示它们之间可以任意流通

代码如下：

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <iostream>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define MAXN 500047#define MAXM 1247int head[MAXM], pre[MAXM];int dep[MAXM], cur[MAXM], gap[MAXM];int EN;int con[MAXM][MAXM];//记录商人与猪栏之间是否相连struct Edge{    int to,next,cap,flow;} edge[MAXN]; //注意是MAXMint tol;int k, c, m;int s, e;//源点，汇点int map[MAXM][MAXM];//加边，单向图三个参数，双向图四个参数void addedge(int u,int v,int w,int rw = 0){    edge[tol].to = v;    edge[tol].cap = w;    edge[tol].flow = 0;    edge[tol].next = head[u];    head[u] = tol++;    edge[tol].to = u;    edge[tol].cap = rw;    edge[tol].flow = 0;    edge[tol].next = head[v];    head[v] = tol++;}int Q[MAXN];void BFS(int start,int end){    memset(dep,-1,sizeof(dep));    memset(gap,0,sizeof(gap));    gap[0] = 1;    int front = 0, rear = 0;    dep[end] = 0;    Q[rear++] = end;    while(front != rear)    {        int u = Q[front++];        for(int i = head[u]; i != -1; i = edge[i].next)        {            int v = edge[i].to;            if(dep[v] != -1)continue;            Q[rear++] = v;            dep[v] = dep[u] + 1;            gap[dep[v]]++;        }    }}int S[MAXN];//输入参数：起点、终点、点的总数//点的编号没有影响，只要输入点的总数int sap(int start,int end,int N){    BFS(start,end);    memcpy(cur,head,sizeof(head));    int top = 0;    int u = start;    int ans = 0;    while(dep[start] < N)    {        if(u == end)        {            int Min = INF;            int inser;            for(int i = 0; i < top; i++)                if(Min > edge[S[i]].cap - edge[S[i]].flow)                {                    Min = edge[S[i]].cap - edge[S[i]].flow;                    inser = i;                }            for(int i = 0; i < top; i++)            {                edge[S[i]].flow += Min;                edge[S[i]^1].flow -= Min;            }            ans += Min;            top = inser;            u = edge[S[top]^1].to;            continue;        }        bool flag = false;        int v;        for(int i = cur[u]; i != -1; i = edge[i].next)        {            v = edge[i].to;            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])            {                flag = true;                cur[u] = i;                break;            }        }        if(flag)        {            S[top++] = cur[u];            u = v;            continue;        }        int Min = N;        for(int i = head[u]; i != -1; i = edge[i].next)            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)            {                Min = dep[edge[i].to];                cur[u] = i;            }        gap[dep[u]]--;        if(!gap[dep[u]])return ans;        dep[u] = Min + 1;        gap[dep[u]]++;        if(u != start)u = edge[S[--top]^1].to;    }    return ans;}void init(){    memset(head,-1,sizeof(head));    memset(con,0,sizeof(con));    EN = 0;}int main(){    int m, n;    while(~scanf("%d%d",&m,&n))    {        init();        int num;        for(int i = 1; i <= m; i++)        {            scanf("%d",&num);            addedge(0, i, num);//源点        }        int a, b, c[MAXM];        for(int i = 1; i <= n; i++)        {            scanf("%d",&a);            for(int j = 1; j <= a; j++)            {                scanf("%d",&c[j]);                addedge(c[j], m+i, INF);//猪和商人                con[i][c[j]] = 1;//第i个商人能买的            }            scanf("%d",&b);//能买几头            addedge(m+i, m+n+1, b);//商人和汇点        }        for(int i = 1; i < n; i++)//商人与商人之间，从前到后        {            int vis[MAXM];//记录是否已经加过边            memset(vis,0,sizeof(vis));            for(int j = 1; j <= m; j++)            {                if (con[i][j])//如果商人能买                {                    for(int k = i+1; k <= n; k++)                    {                        if (con[k][j] && !vis[k])//判断商人k和猪栏j是否相连                        {                            addedge(m+i, m+k, INF);                            vis[k] = 1;                        }                    }                }            }        }        int sta = 0;//起点        int end = m+n+1;//汇点        int N = m+n+2;//总结点数        int ans = sap(sta,end,N);        printf("%d\n",ans);    }    return 0;}