poj1149PIGS(最大流+建图)

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PIGS

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 16172 Accepted: 7249

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

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题目链接：

http://poj.org/problem?id=1149

题意：M个猪圈，N个顾客，每个顾客有一些的猪圈的钥匙，只能购买能打开的猪圈里的猪，而且要买一定数量的猪，每个猪圈有已知数量的猪，但是猪圈可以重新打开，将猪的个数，重新分配，但是只能将猪往当前打开状态的猪圈里赶，以达到卖出的猪的数量最多。

 

思路             ：还是4部分，源点->猪圈->猪圈->汇点      

Accepted        976K63MSC++

     

能用EK水的，当然用EK水    

此题需要建图

代码如下：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <math.h>#include <queue>#define init(a) memset(a,0,sizeof(a))#define PI acos(-1,0)using namespace std;const int maxn = 500;const int maxm = 40000;#define lson left, m, id<<1#define rson m+1, right, id<<1|1#define min(a,b) (a>b)?b:a#define max(a,b) (a>b)?a:b#define MAX INT_MAX    //忽视这些头文件即可int c[1000][1000];int re[1000];int f[1000][1000];int p[1000];int n,m;void initt()//初始化操作{    int i,j;    for(i=0;i<=m+1;i++)    {        for(j=0;j<=m+1;j++)        {            c[i][j]=f[i][j]=0;        }        p[i]=0;    }}void EK(int s,int t)//此题用EK算法水过{    queue<int >q;    while(!q.empty())        q.pop();    int sum=0;    while(1)//放入一个死循环中    {        memset(re,0,sizeof(re));        q.push(s);        re[s]=MAX;        p[s]=-1;        while(!q.empty())//BFS找到增广路（增广链）        {            int u=q.front();            q.pop();            for(int i=1;i<=m+1;i++)//找到新节点i            {                if(!re[i]&&f[u][i]<c[u][i])                {                    q.push(i);                    p[i]=u;//记录i的父亲并且加入队列                    re[i]=min(re[u],c[u][i]-f[u][i]);//找到s-t的最小残余流量                }            }        }        if(re[t]==0) break;//找不到，当前已经是最大流        for(int st=t; st!=s ;st=p[st])//从汇点往回走        {            f[p[st]][st]+=re[t];//更新正向流量            f[st][p[st]]-=re[t];//更新反向流量        }        sum+=re[t];//更新从s流出的总流量    }    printf("%d\n",sum);//打印出总的流量}int main(){    int pig[1001];    int a,b,w;    int i,j;    while(~scanf("%d %d",&n,&m))    {        initt();        for(i=1;i<=n;i++)            scanf("%d",&pig[i]);        for(i=1;i<=m;i++)        {            scanf("%d",&a);            while(a--)            {                scanf("%d",&b);                if(!p[b])//判断当前猪圈是否打开过                {                    c[p[b]][i]+=pig[b];                    p[b]=i;                }                else                {                    c[p[b]][i]=MAX;//可以从其他猪圈流向本猪圈，流量可能为无限大                    p[b]=i;                }            }            scanf("%d",&w);            c[i][m+1]+=w;        }        EK(0,m+1);    }    return 0;}