poj 1149 PIGS 【网络最大流】

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 16630		Accepted: 7457

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

分析：此题关键和难点是建图，建立源点和汇点，把顾客当作源点和汇点之外的节点。使每个打开猪笼第一个

顾客与源点链接，权值为此猪笼的猪的数量，其他顾客与其要打开的猪笼的上以为顾客链接权值为INF。

最后把每位顾客与汇点相连，权值为次顾客索要猪的数量。这样图就建好了，跑一便最大流即可。

代码示例：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>
#define Lh 2000
#define Le 200000
#define max 1000000000
using namespace std;
typedef struct
{
int to;
int w;
int next;
}node;
typedef struct
{
int x;
int t;
}DEP;
node E[Le];
DEP fir,nex;
int head[Lh],headx[Lh],deep[Lh],cnt;
void ADD(int a,int b,int c)
{
E[++cnt].to=b;
E[cnt].w=c;
E[cnt].next=head[a];
head[a]=cnt;
E[++cnt].to=a;
E[cnt].w=0;
E[cnt].next=head[b];
head[b]=cnt;
}
int min(int x,int y)
{
return x<y?x:y;
}
int bfs(int s,int t,int n)
{
memset(deep,255,sizeof(deep));
queue<DEP>Q;
fir.x=s;
fir.t=0;
deep[s]=0;
Q.push(fir);
while(!Q.empty())
{
fir=Q.front();
Q.pop();
for(int i=head[fir.x];i;i=E[i].next)
{
nex.x=E[i].to;
nex.t=fir.t+1;
if(deep[nex.x]!=-1||!E[i].w)
continue;
deep[nex.x]=nex.t;
Q.push(nex);
}
}
for(int i=0;i<=n;i++)
headx[i]=head[i];
return deep[t]!=-1;
}
int dfs(int s,int t,int flow)
{
if(s==t)
return flow;
int newflow=0;
for(int i=headx[s];i;i=E[i].next)
{
headx[s]=i;
int to=E[i].to;
int w=E[i].w;
if(!w||deep[to]!=deep[s]+1)
continue;
int temp=dfs(to,t,min(w,flow-newflow));
newflow+=temp;
E[i].w-=temp;
E[i^1].w+=temp;
if(newflow==flow)
break;
}
if(!newflow)deep[s]=0;
return newflow;
}
int Dinic(int s,int t,int m)
{
int sum=0;
while(bfs(s,t,m))
{
sum+=dfs(s,t,max);
}
return sum;
}
int main()
{
int m,n;
int pig[1010],pre[1010];
int a,k,b;
int s,t,num;
while(~scanf("%d%d",&m,&n))
{
memset(head,0,sizeof(head));
cnt=1;
memset(pre,0,sizeof(pre));
for(int i=1;i<=m;i++)
scanf("%d",&pig[i]);
s=n+1;
t=n+2;
num=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a);
for(int j=1;j<=a;j++)
{
scanf("%d",&k);
if(!pre[k])
{
ADD(s,i,pig[k]);
num++;
}
else
{
ADD(pre[k],i,max);
num++;
}
pre[k]=i;
}
scanf("%d",&b);
ADD(i,t,b);
num++;
}
printf("%d\n",Dinic(s,t,num));
}
return 0;
}