POJ 1149 网络流最大流 解题报告

PIGS

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

【解题报告】(转载自网络流建模汇by Edelweiss )













代码如下:

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
using namespace std;
#define inf 0x3f3f3f3f
#define maxv 10005
#define maxe 100005

int nume=0,head[maxv],e[maxe][3];

void inline adde(int i,int j,int c)
{
e[nume][0]=j;e[nume][1]=head[i];head[i]=nume;
e[nume++][2]=c;
e[nume][0]=i;e[nume][1]=head[j];head[j]=nume;
e[nume++][2]=0;
}

int ss,tt,n,m;
int vis[maxv],lev[maxv];

bool bfs()
{
for(int i=0;i<maxv;i++)
vis[i]=lev[i]=0;
queue<int>q;
q.push(ss);
vis[ss]=1;
while(!q.empty())
{
int cur=q.front();
q.pop();
for(int i=head[cur];i!=-1;i=e[i][1])
{
int v=e[i][0];
if(!vis[v]&&e[i][2]>0)
{
lev[v]=lev[cur]+1;
vis[v]=1;
q.push(v);
}
}
}
return vis[tt];
}
int dfs(int u,int minf)
{
if(u==tt||minf==0)return minf;
int sumf=0,f;
for(int i=head[u];i!=-1&&minf;i=e[i][1])
{
int v=e[i][0];
if(lev[v]==lev[u]+1&&e[i][2]>0)
{
f=dfs(v,minf<e[i][2]?minf:e[i][2]);
e[i][2]-=f;e[i^1][2]+=f;
sumf+=f;minf-=f;
}
}
if(!sumf) lev[u]=-1;
return sumf;
}
int Dinic()
{
int sum=0;
while(bfs())sum+=dfs(ss,inf);
return sum;
}

vector<int>a[200];
int pig[1005];
int L[1005];

int main()
{
memset(head,-1,sizeof(head));
scanf("%d%d",&m,&n);
ss=0,tt=n+1;
for(int i=1;i<=m;++i) scanf("%d",&pig[i]);
for(int i=1;i<=n;++i)
{
int x,y;
scanf("%d",&x);
while(x--)
{
scanf("%d",&y);
a[i].push_back(y);
}
scanf("%d",&y);
adde(i,tt,y);
}
for(int i=1;i<=n;i++)
for(int j=0;j<a[i].size();j++)
{
int v=a[i][j];
if(!L[v])
{
L[v]=i;
adde(ss,i,pig[v]);
}
else
{
adde(L[v],i,inf);
L[v]=i;
}
}
printf("%d\n",Dinic());
return 0;
}

让我看到你们的双手