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 程式師世界 >> 編程語言 >> C語言 >> C++ >> 關於C++ >> poj1149 PIGS

poj1149 PIGS

編輯:關於C++

 

PIGS Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 19068   Accepted: 8697

 

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

Source

Croatia OI 2002 Final Exam - First day

 

 

 

最大流。

構圖方式:

①把每個顧客看作除源點和匯點以外的節點。

②從源點向每個豬圈的第一個顧客連一條邊,容量為該豬圈最初的豬的數量。

③每個豬圈的前後兩個顧客之間連一條邊,容量為正無窮。因為可以任意分配每個豬圈中的豬的數量。

④從每個顧客向匯點連一條邊,容量為要購買的豬的數量。

這道題的構圖方法很巧妙。


 

 

 

#include
#include
#include
#include
#include
#include
#include
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define LL long long
#define pa pair
#define MAXN 105
#define MAXM 1005
#define INF 1000000000
using namespace std;
int n,m,k,x,s,t,cnt=1,ans=0;
int pre[MAXM],head[MAXN],cur[MAXN],dis[MAXN],c[MAXN],a[MAXM];
bool vst[MAXM],f[MAXN];
struct edge_type
{
	int next,to,v;
}e[10005];
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
inline void add_edge(int x,int y,int v)
{
	e[++cnt]=(edge_type){head[x],y,v};head[x]=cnt;
	e[++cnt]=(edge_type){head[y],x,0};head[y]=cnt;
}
inline bool bfs()
{
	queueq;
	memset(dis,-1,sizeof(dis));
	dis[s]=0;q.push(s);
	while (!q.empty())
	{
		int tmp=q.front();q.pop();
		if (tmp==t) return true;
		for(int i=head[tmp];i;i=e[i].next) if (e[i].v&&dis[e[i].to]==-1)
		{
			dis[e[i].to]=dis[tmp]+1;
			q.push(e[i].to);
		}
	}
	return false;
}
inline int dfs(int x,int f)
{
	int tmp,sum=0;
	if (x==t) return f;
	for(int &i=cur[x];i;i=e[i].next)
	{
		int y=e[i].to;
		if (e[i].v&&dis[y]==dis[x]+1)
		{
			tmp=dfs(y,min(f-sum,e[i].v));
			e[i].v-=tmp;e[i^1].v+=tmp;sum+=tmp;
			if (sum==f) return sum;
		}
	}
	if (!sum) dis[x]=-1;
	return sum;
}
inline void dinic()
{
	while (bfs())
	{
		F(i,1,n+2) cur[i]=head[i];
		ans+=dfs(s,INF);
	}
}
int main()
{
	memset(head,0,sizeof(head));
	memset(c,0,sizeof(c));
	memset(vst,false,sizeof(vst));
	m=read();n=read();s=n+1;t=n+2;
	F(i,1,m) a[i]=read();
	F(i,1,n)
	{
		memset(f,false,sizeof(f));
		k=read();
		while (k--)
		{
			x=read();
			if (!vst[x]) {vst[x]=true;c[i]+=a[x];}
			if (pre[x]&&!f[pre[x]]) add_edge(pre[x],i,INF),f[pre[x]]=true;
			pre[x]=i;
		} 
		x=read();if (x) add_edge(i,t,x);
	}
	F(i,1,n) if (c[i]) add_edge(s,i,c[i]);
	dinic();
	printf("%d\n",ans);
}


 

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