PIGS
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 13677 | Accepted: 6044 |
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.
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.
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
这题难点在于如何建图，如何理解题目中的“if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses.”也将成为建图的关键。
如果我们把每个顾客作为图中的点，则对于拥有某个猪圈的钥匙的连续的两个顾客i和j,建一条边i->j。现在解释为何可以这样建边：比如说i拥有猪圈1,2，3的钥匙，j拥有猪圈1的钥匙，这样i,j同时拥有3号猪圈的钥匙，按理说j只能打开1号猪圈，但i顾客购买猪后，Mirko是可以随意分配1,2,3中猪的数量的，所以j顾客实质上是可以得到1,2,3中的猪的。
因为猪圈有m个，所以我们可以加一个源点s，分别与n个顾客建边，鉴于上面的建边方式，s应与每个猪圈的第一个顾客建边，边容量为猪圈中猪的数量，若某个顾客同时为多个猪圈的第一个顾客，则数量相加。
每个顾客购买猪的数量是有上限的，所以我们应该引入一个汇点t，每个顾客与t建一条边，边容量为每个顾客需要购买的猪的数量。
#include <iostream>
#include<cstdio>
using namespace std;
const int MAXN=105;
const int INF=(1<<29);
int flow[MAXN][MAXN];//容量限制
int dalta[MAXN];//改变量
int pre[MAXN];
int flag[MAXN];//标号
int m,n;
int EK()
{
int i,maxflow=0;
int queue[MAXN],front,rear;
while(1)
{
front=rear=0;
for(i=1;i<=n+1;i++)
flag[i]=0;
flag[0]=1;
pre[0]=0;
dalta[0]=INF;
queue[rear++]=0;
while(front!=rear&&!flag[n+1])
{
int v=queue[front++];
for(i=1;i<=n+1;i++)
{
if(flag[i])
continue;
if(flow[v][i])
{
dalta[i]=min(dalta[v],flow[v][i]);
flag[i]=1;
pre[i]=v;
queue[rear++]=i;
}
}
}
if(!flag[n+1])
break;
maxflow+=dalta[n+1];
i=n+1;
while(i!=0)
{
flow[pre[i]][i]-=dalta[n+1];
flow[i][pre[i]]+=dalta[n+1];
i=pre[i];
}
}
return maxflow;
}
int main()
{
int i,j;
int pigs[1005];//每个猪圈中的猪的数量
int before[1005];//每个猪圈的前一个顾客
while(~scanf("%d%d",&m,&n))
{
for(i=0;i<=n+1;i++)
for(j=0;j<=n+1;j++)
flow[i][j]=0;
for(i=1;i<=m;i++)
{
scanf("%d",pigs+i);
before[i]=-1;
}
for(i=1;i<=n;i++)
{
int num;
scanf("%d",&num);
while(num--)
{
scanf("%d",&j);
if(before[j]==-1)
{
before[j]=i;
flow[0][i]+=pigs[j];
}
else
{
flow[before[j]][i]=INF;
before[j]=i;
}
}
scanf("%d",&j);
flow[i][n+1]=j;
}
printf("%d\n",EK());
}
return 0;
}