POJ3281：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 prepared D (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 Fi and Di, the number of dishes that cow i likes and the number of drinks that cow i likes. The next Fi integers denote the dishes that cow i will eat, and the Di 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.

题目大意
农夫约翰为他的N头牛准备了F种食物和D种饮料。每头牛都有各自喜欢的食物和饮料，而每种食物或饮料只能分配给一头牛。最多能有多少头牛可以同时得到喜欢的食物和饮料？

解题思路
可以转化为最大流问题。
将食物和饮料所对应的两个匹配通过下面的方法联合起来求解
（1）图的顶点在食物对应的匹配中的食物和牛，饮料对应的匹配中的饮料和牛之外，还有一个源点s和一个汇点t。
（2）在两个匹配相同的牛之间连一条边，在s和所有食物，t和所有饮料之间连一条边。
（3）边的方向为s->食物->牛->牛->饮料->t，容量全都为1。
这个图中的每一条s-t路径都对应一个牛的食物和饮料的分配方案。我们把食物所对应的牛和饮料所对应的牛拆成两个顶点，之间连一条容量为1的边，就保证了一头牛不会被分配多组食物和饮料。
所以，问题转化为计算该图中的最大流。
代码如下：

#include<iostream>#include<vector>#include<algorithm>using namespace std;struct edge{int to,cap,rev;};vector<edge> G[1000];bool used[1000];int N,F,D;bool likeF[105][105];bool likeD[105][105];void add_edge(int from,int to,int cap){    edge temp;    temp.to=to;    temp.cap=cap;    temp.rev=G[to].size();    G[from].push_back(temp);    temp.to=from;    temp.cap=0;    temp.rev=G[from].size()-1;    G[to].push_back(temp);}int dfs(int v,int t,int f){    if(v==t)    return f;    used[v]=true;    for(int i=0;i<G[v].size();i++)    {        edge &e=G[v][i];        if(!used[e.to]&&e.cap>0)        {            int d=dfs(e.to,t,min(f,e.cap));            if(d>0)            {                e.cap-=d;                G[e.to][e.rev].cap+=d;                return d;            }        }    }    return 0;}int max_flow(int s,int t){    int flow=0;    for(;;)    {        memset(used,0,sizeof(used));        int f=dfs(s,t,10000000);        if(f==0)        return flow;        flow+=f;    }}void solve(){    int s=2*N+F+D;    int t=s+1;    for(int i=0;i<F;i++)    {       add_edge(s,2*N+i,1);    }    for(int i=0;i<D;i++)    {        add_edge(2*N+F+i,t,1);    }    for(int i=0;i<N;i++)    {        add_edge(i,N+i,1);    }    for(int i=0;i<N;i++)    {      for(int j=0;j<F;j++)      {        if(likeF[i][j])        add_edge(2*N+j,i,1);      }      for(int j=0;j<D;j++)      {        if(likeD[i][j])        add_edge(N+i,2*N+F+j,1);      }    }    cout<<max_flow(s,t)<<endl;}int main(){    int i,j,f,d,x;    while(cin>>N>>F>>D)    {        memset(likeF,0,sizeof(likeF));        memset(likeD,0,sizeof(likeD));        for(i=0;i<N;i++)        {            cin>>f>>d;            while(f--)            {                cin>>x;                likeF[i][x-1]=1;            }            while(d--)            {                cin>>x;                likeD[i][x-1]=1;            }        }        solve();    }    return 0;}