poj1149 pigs 網絡流

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17680		Accepted: 8015

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 3
3 1 10
2 1 2 2
2 1 3 3
1 2 6

Sample Output

7

題意：略；

以顾客为点，加一个源点和汇点，源点到顾客的权为他能去的猪槽的猪数量总和，顾客到汇点的权为顾客可以买的朱数量。顾客间如果有相同的猪槽加一条INF的边。

#include <iostream>
#include <stdio.h>
#include <queue>
#include <cmath>
#include <string.h>
#include <vector>
#include <algorithm>
#include <map>
#include <queue>
using namespace std;

#define LL                                      long long
#define scan(a)                             scanf("%d",&a)
#define maxn                                1111
#define REP(i,a,b)                          for(int i=a;i<b;++i)
#define mset(a,b)                           memset(a,b,sizeof a)

const LL mod = 1000000000;

const int MAXN = 111;//点数的最大值
const int MAXM = 11100;//边数的最大值
const int INF = 0x3f3f3f3f;
int pigs[1111];
vector<int> pigRoom[1111];
struct Edge
{
    int to,next,cap,flow;
}edge[MAXM];//注意是MAXM

int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];

void init()
{
    tol = 0;
    memset(head,-1,sizeof(head));
}
//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].next = head[u];
    edge[tol].flow= 0;head[u] = tol++;
    edge[tol].to =u;edge[tol].cap = rw;edge[tol].next = head[v];
    edge[tol].flow= 0;head[v]=tol++;
}
//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)
{
    memset(gap,0,sizeof(gap));
    memset(dep,0,sizeof(dep));
    memcpy(cur,head,sizeof(head));
    int u = start;
    pre[u] = -1;
    gap[0] = N;
    int ans = 0;
    while(dep[start] < N)
    {
        if(u == end)
        {
            int Min = INF;
            for(int i = pre[u];i != -1; i = pre[edge[i^1].to])
                if(Min > edge[i].cap - edge[i].flow)
                    Min = edge[i].cap - edge[i].flow;
            for(int i = pre[u];i != -1; i = pre[edge[i^1].to])
            {
                edge[i].flow += Min;
                edge[i^1].flow -= Min;
            }
            u = start;
            ans += Min;
            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] = pre[v] = i;
                break;
            }
        }
        if(flag)
        {
            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[pre[u]^1].to;
    }
    return ans;
}

int main()
{
    int s=0;
    int n,m;
    while(cin>>m>>n)
    {
        //mset(minf,0);
        int e=n+1;
        init();
        REP(i,1,m+1)  scan(pigs[i]);
        REP(i,1,n+1)
        {
            int A,sump=0,B;
            scan(A);
            REP(j,0,A)
            {
                int room;
                scan(room);
                if(pigRoom[room].size()==0)
                    sump+=pigs[room];
                pigRoom[room].push_back(i);
            }
            scan(B);
            addedge(s,i,sump);
            addedge(i,e,B);
        }
        REP(i,1,m+1)
        {
            int sz=pigRoom[i].size();
            for(int j=0;j<sz-1;++j)
            {
                for(int k=j+1;k<sz;++k)
                {
                    addedge(pigRoom[i][j],pigRoom[i][k],INF);
                }
            }
        }
        printf("%d\n",sap(s,e,n+2));
    }
}