POJ1149 PIGS

PIGS

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 21349 Accepted: 9755

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

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题目大意：有N个顾客，有M个猪圈，每个猪圈有一定的猪，在开始的时候猪圈都是关闭的，顾客来买猪，顾客打开某个猪圈，可以在其中挑选一定的猪的数量，在这个顾客走后，可以在打开的猪圈中将某个猪圈的一些猪牵到另外一个打开的猪圈，然后所有的猪圈会关闭，这样下一个顾客来了继续上面的工作

现在求怎么一种方法，使所有顾客买的猪的总数最大

思路：猪圈数太大了，建边其实可以省略，直接把它的猪全部给第一个遇到的人，在记录下它的归属即可。建图源点和每个猪圈的第一个顾客连边，边的权是开始时猪圈中猪的数目，顾客j紧跟在顾客i之后打开某个猪圈，则边<i,j>的权是无限大每个顾客和汇点之间连边，边的权是顾客所希望购买的猪的数目

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;#define MAXN 500struct node{    int u, v, next, cap;} edge[MAXN*MAXN];int nt[MAXN], s[MAXN], d[MAXN], visit[MAXN];int cnt;int a[2005];int b[2005];void init(){    cnt = 0;    memset(s, -1, sizeof(s));}void add(int u, int v, int c){    edge[cnt].u = u;    edge[cnt].v = v;    edge[cnt].cap = c;    edge[cnt].next = s[u];    s[u] = cnt++;    edge[cnt].u = v;    edge[cnt].v = u;    edge[cnt].cap = 0;    edge[cnt].next = s[v];    s[v] = cnt++;}bool BFS(int ss, int ee){    memset(d, 0, sizeof d);    d[ss] = 1;    queue<int>q;    q.push(ss);    while (!q.empty())    {        int pre = q.front();        q.pop();        for (int i = s[pre]; ~i; i = edge[i].next)        {            int v = edge[i].v;            if (edge[i].cap > 0 && !d[v])            {                d[v] = d[pre] + 1;                q.push(v);            }        }    }    return d[ee];}int DFS(int x, int exp, int ee){    if (x == ee||!exp) return exp;    int temp,flow=0;    for (int i = nt[x]; ~i ; i = edge[i].next, nt[x] = i)    {        int v = edge[i].v;        if (d[v] == d[x] + 1&&(temp = (DFS(v, min(exp, edge[i].cap), ee))) > 0)        {            edge[i].cap -= temp;            edge[i ^ 1].cap += temp;            flow += temp;            exp -= temp;            if (!exp) break;        }    }    if (!flow) d[x] = 0;    return flow;}int Dinic_flow(int ss, int ee){    int ans = 0;    while (BFS(ss, ee))    {        for (int i = 0; i <= ee; i++) nt[i] = s[i];        ans+= DFS(ss, INF, ee);    }    return ans;}int main(){    int n,m,u,v,c;    while (~scanf("%d %d", &m, &n))    {        init();        for(int i=1;i<=m;i++)            scanf("%d",&a[i]);        memset(b,0,sizeof b);        for(int i=1; i<=n; i++)        {            scanf("%d",&m);            for(int j=0;j<m;j++)            {                scanf("%d",&u);                if(!b[u])                    add(0,i,a[u]),b[u]=i;                else                    add(b[u],i,INF);            }            scanf("%d",&u);            add(i,n+1,u);        }        printf("%d\n",Dinic_flow(0,n+1));    }    return 0;}