HDU 4292 Food （网络流，最大流）

HDU 4292 Food （网络流，最大流）

Description

You, a part-time dining service worker in your college’s dining hall, are now confused with a new problem: serve as many people as possible.

The issue comes up as people in your college are more and more difficult to serve with meal: They eat only some certain kinds of food and drink, and with requirement unsatisfied, go away directly.

You have prepared F (1 <= F <= 200) kinds of food and D (1 <= D <= 200) kinds of drink. Each kind of food or drink has certain amount, that is, how many people could this food or drink serve. Besides, You know there’re N (1 <= N <= 200) people and you too can tell people’s personal preference for food and drink.

Back to your goal: to serve as many people as possible. So you must decide a plan where some people are served while requirements of the rest of them are unmet. You should notice that, when one’s requirement is unmet, he/she would just go away, refusing any service.

Input

There are several test cases.

For each test case, the first line contains three numbers: N,F,D, denoting the number of people, food, and drink.

The second line contains F integers, the ith number of which denotes amount of representative food.

The third line contains D integers, the ith number of which denotes amount of representative drink.

Following is N line, each consisting of a string of length F. e jth character in the ith one of these lines denotes whether people i would accept food j. “Y” for yes and “N” for no.

Following is N line, each consisting of a string of length D. e jth character in the ith one of these lines denotes whether people i would accept drink j. “Y” for yes and “N” for no.

Please process until EOF (End Of File).

Output

For each test case, please print a single line with one integer, the maximum number of people to be satisfied.

Sample Input

4 3 3

1 1 1

1 1 1

YYN

NYY

YNY

YNY

YNY

YYN

YYN

NNY

Sample Output

3

Http

HDU:https://vjudge.net/problem/HDU-4292

Hit

网络流，最大流

题目大意

现在提供若干种饮料、食物，每种都有一定的数量。给定N个人对每一种食物和饮料的喜好，若给某人提供了其喜欢的食物和饮料，则称这个人是开心的。现在求能使最多的人开心的数量。

解决思路

我们按照源点-食物-人-饮料-汇点的顺序建图。对于每一个食物，从源点连容量为其数量的的边，而对于每一个人，连接他和他所有喜欢的食物，容量都是1,。因为给一个人只提供一份食物和一份饮料，所以我们把人拆点，中间连容量为1的边。而对于饮料则是类似的，从人到他喜欢的饮料连容量为1的边，从饮料到汇点连容量为其数量的边。这样跑最大流即可。

关于最大流，这里使用Dinic，可以参考这篇文章

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxN=1001;
const int maxM=maxN*maxN*2;
const int inf=2147483647;

class Edge
{
public:
	int u,v,flow;
};

int N,F,D;//点的编号安排：0汇点，[1,N]人1，[N+1,N*2]人2，[N*2+1,N*2+F]食物，[N*2+F+1,N*2+F+D]饮料，N*2+F+D+1汇点
int cnt=-1;
char str[maxN];
int Head[maxN];
int Next[maxM];
Edge E[maxM];
int depth[maxN];
int cur[maxN];
int Q[maxN];

void Add_Edge(int u,int v,int flow);
bool bfs();
int dfs(int u,int flow);

int main()
{
	while (cin>>N>>F>>D)//多组数据
	{
		cnt=-1;//先清空
		memset(Head,-1,sizeof(Head));
		for (int i=1;i<=F;i++)//读入食物的数量
		{
			int maxflow;
			scanf("%d",&maxflow);
			Add_Edge(0,N*2+i,maxflow);//连接源点和食物
		}
		for (int i=1;i<=D;i++)//读入饮料的数量
		{
			int maxflow;
			scanf("%d",&maxflow);
			Add_Edge(N*2+F+i,N*2+F+D+1,maxflow);//连接饮料和汇点
		}
		for (int i=1;i<=N;i++)
		{
			scanf("%s",str);
			for (int j=0;j<F;j++)
				if (str[j]=='Y')
					Add_Edge(N*2+j+1,i,1);//连接人与食物
		}
		for (int i=1;i<=N;i++)
		{
			scanf("%s",str);
			for (int j=0;j<D;j++)
				if (str[j]=='Y')
					Add_Edge(N+i,N*2+F+j+1,1);//连接人与饮料
		}
		for (int i=1;i<=N;i++)
			Add_Edge(i,N+i,1);//连接人1与人2，即拆开的两个点
		int Ans=0;//求解最大流
		while (bfs())
		{
			for (int i=0;i<=2*N+F+D+1;i++)
				cur[i]=Head[i];
			while (int di=dfs(0,inf))
				Ans+=di;
		}
		cout<<Ans<<endl;
	}
	return 0;
}

void Add_Edge(int u,int v,int flow)
{
	cnt++;
	Next[cnt]=Head[u];
	Head[u]=cnt;
	E[cnt].u=u;
	E[cnt].v=v;
	E[cnt].flow=flow;

	cnt++;
	Next[cnt]=Head[v];
	Head[v]=cnt;
	E[cnt].u=v;
	E[cnt].v=u;
	E[cnt].flow=0;
}

bool bfs()
{
	memset(depth,-1,sizeof(depth));
	int h=1,t=0;
	depth[0]=1;
	Q[1]=0;
	do
	{
		t++;
		int u=Q[t];
		for (int i=Head[u];i!=-1;i=Next[i])
		{
			int v=E[i].v;
			if ((depth[v]==-1)&&(E[i].flow>0))
			{
				depth[v]=depth[u]+1;
				h++;
				Q[h]=v;
			}
		}
	}
	while (t!=h);
	if (depth[N*2+F+D+1]==-1)
		return 0;
	return 1;
}

int dfs(int u,int flow)
{
	if (u==N*2+F+D+1)
		return flow;
	for (int i=Head[u];i!=-1;i=Next[i])
	{
		int v=E[i].v;
		if ((depth[v]==depth[u]+1)&&(E[i].flow>0))
		{
			int di=dfs(v,min(flow,E[i].flow));
			if (di>0)
			{
				E[i].flow-=di;
				E[i^1].flow+=di;
				return di;
			}
		}
	}
	return 0;
}