数量queuepoj1149 PIGS
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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.
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.
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，如何懂得题目中的“if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses.”也将成为建图的症结。
如果我们把每个主顾作为图中的点，则对于具有某个猪圈的钥匙的连续的两个主顾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具有猪圈1,2，3的钥匙，j具有猪圈1的钥匙，这样i,j同时具有3号猪圈的钥匙，按理说j只能打开1号猪圈，但i主顾购买猪后，Mirko是可以随便分配1,2,3中猪的数量的，所以j主顾实质上是可以得到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，分离与n个主顾建边，鉴于下面的建边方式，s应与每个猪圈的第一个主顾建边，边容量为猪圈中猪的数量，若某个主顾同时为多个猪圈的第一个主顾，则数量相加。
每个主顾购买猪的数量是有上限的，所以我们应该引入一个汇点t，每个主顾与t建一条边，边容量为每个主顾须要购买的猪的数量。
#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];//每个猪圈的前一个主顾
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;
}
文章结束给大家分享下程序员的一些笑话语录： 打赌
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数量和queue
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