poj Dining 3281 （最大流问题&&机智建图） 好题

Dining

Description

Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others.

Farmer John has cooked fabulous meals for his cows, but he forgot to check his menu against their preferences. Although he might not be able to stuff everybody, he wants to give a complete meal of both food and drink to as many cows as possible.

Farmer John has cooked F (1 ≤ F ≤ 100) types of foods and preparedD (1 ≤ D ≤ 100) types of drinks. Each of his N (1 ≤ N ≤ 100) cows has decided whether she is willing to eat a particular food or drink a particular drink. Farmer John must assign a food type and a drink type to each cow to maximize the number of cows who get both.

Each dish or drink can only be consumed by one cow (i.e., once food type 2 is assigned to a cow, no other cow can be assigned food type 2).

Input

Line 1: Three space-separated integers: N, F, and D
Lines 2.. N+1: Each line i starts with a two integers F_i and D_i, the number of dishes that cow i likes and the number of drinks that cow i likes. The next F_i integers denote the dishes that cow i will eat, and the D_i integers following that denote the drinks that cow i will drink.

Output

Line 1: A single integer that is the maximum number of cows that can be fed both food and drink that conform to their wishes

Sample Input

4 3 3
2 2 1 2 3 1
2 2 2 3 1 2
2 2 1 3 1 2
2 1 1 3 3

Sample Output

3

Hint

One way to satisfy three cows is:
Cow 1: no meal
Cow 2: Food #2, Drink #2
Cow 3: Food #1, Drink #1
Cow 4: Food #3, Drink #3
The pigeon-hole principle tells us we can do no better since there are only three kinds of food or drink. Other test data sets are more challenging, of course.

//题意：

有n头牛，f种食物，d种饮料

下面是n行输入，每行表示一头牛的喜好情况（第i行表示第i头牛的喜好）。

每行输入先输入两个数字，ff,dd。分别表示这头牛有ff种喜爱的食物，dd种喜爱的饮料。接着后面分别输入ff种食物的代号，dd种饮料的代号

问最多能满足几头牛的喜好（都吃上自己喜爱的食物和饮料）

//思路：

添加总汇点，和总节点，让其变成一个最大流的图。

具体看代码。。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#define INF 0x3f3f3f3f
#define N 410
using namespace std;
int dis[N];
int vis[N];
int c[N];
int head[N],edgenum;
struct zz
{
	int from;
	int to;
	int cap;
	int flow;
	int next;
}edge[N<<4];
void add(int u,int v,int w)
{
	zz E={u,v,w,0,head[u]};
	edge[edgenum]=E;
	head[u]=edgenum++;
	zz EE={v,u,0,0,head[v]};
	edge[edgenum]=EE;
	head[v]=edgenum++;
}
int n,f,d;
bool bfs(int s,int e)
{
	queue<int>q;
	memset(dis,-1,sizeof(dis));
	memset(vis,0,sizeof(vis));	
	while(!q.empty())
		q.pop();
	q.push(s);
	dis[s]=0;
	vis[s]=1;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		for(int i=head[u];i!=-1;i=edge[i].next)
		{
			zz v=edge[i];
			if(!vis[v.to]&&v.cap>v.flow)
			{
				vis[v.to]=1;
				dis[v.to]=dis[u]+1;
				if(v.to==e)
					return true;
				q.push(v.to);
			}
		}
	}
	return false;
}
int dfs(int x,int a,int e)
{
	if(x==e||a==0)
		return a;
	int flow=0,f;
	for(int i=c[x];i!=-1;i=edge[i].next)
	{
		zz &v=edge[i];
		if(dis[v.to]==dis[x]+1&&(f=dfs(v.to,min(a,v.cap-v.flow),e))>0)
		{
			v.flow+=f;
			edge[i^1].flow-=f;
			flow+=f;
			a-=f;
			if(a==0)
				break;
		}
	}
	return flow;
}
int maxflow(int s,int e)
{
	int flow=0;
	while(bfs(s,e))
	{
		memcpy(c,head,sizeof(head));
		flow+=dfs(s,INF,e);
	}
	return flow;
}
int main()
{
	int ff,dd,i,j;
	while(scanf("%d%d%d",&n,&f,&d)!=EOF)
	{
		edgenum=0;
		memset(head,-1,sizeof(head));
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&ff,&dd);
			while(ff--)
			{
				scanf("%d",&j);
				add(n*2+j,i,1);
			}
			while(dd--)
			{
				scanf("%d",&j);
				add(n+i,n*2+f+j,1);
			}
			add(i,n+i,1);
		}
		for(i=1;i<=f;i++)
			add(0,n*2+i,1);//添加总汇点0 
		for(i=1;i<=d;i++)
			add(n*2+f+i,n*2+f+d+1,1);//添加总结点 n*2+f+d+1
		printf("%d\n",maxflow(0,n*2+f+d+1));
	}
	return 0;
}

//另一种算法

<pre class="cpp" name="code">#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
#define mem(x,y) memset(x,y,sizeof(x))
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=500;
const int MAXM=50010<<1;
int vis[MAXN],map[MAXN][MAXN],pre[MAXN];
bool bfs(int s,int e)
{
	queue<int>dl;
	mem(vis,0);
	mem(pre,-1);
	vis[s]=1;
	dl.push(s);	
	int a;
	while(!dl.empty())
	{
		a=dl.front();
		dl.pop(); 
		if(a==e)
			return true;
		for(int i=1;i<=e;i++)
		{
			if(!vis[i]&&map[a][i])
			{
				dl.push(i);
				vis[i]=1;
				pre[i]=a;				
			}
		}
	}
	return false;
}
int maxflow(int s,int e)
{
	int flow=0;
	while(bfs(s,e))
	{
		int temp=INF;
		int r=e;
		while(r!=s)
			temp=min(temp,map[pre[r]][r]),r=pre[r];
		r=e;
		while(r!=s)
			map[pre[r]][r]-=temp,map[r][pre[r]]+=temp,r=pre[r];
		flow+=temp;
	}
	return flow;
}
int main(){
	int n,F,D,f,d,a,b;
	while(~scanf("%d%d%d",&n,&F,&D))
	{
		mem(map,0);
		for(int i=1;i<=n;i++)
		{
			scanf("%d%d",&f,&d);
			while(f--)
			{
				scanf("%d",&a);
				map[2*n+a][i]=1;
			}
			while(d--)
			{
				scanf("%d",&a);
				map[n+i][2*n+F+a]=1;
			}
			map[i][n+i]=1;
		}
		for(int i=1;i<=F;i++)
			map[0][2*n+i]=1;
		for(int i=1;i<=D;i++)
			map[2*n+F+i][2*n+F+D+1]=1;
		printf("%d\n",maxflow(0,2*n+F+D+1)); 
	}
	return 0;
}