# 洛谷P2891 [USACO07OPEN]吃饭Dining

2018-02-03 19:48:14来源:cnblogs.com作者:自为风月马前卒人点击

## 题目描述

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).

## 输入输出格式

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.

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

## 输入输出样例

`4 3 32 2 1 2 3 12 2 2 3 1 22 2 1 3 1 22 1 1 3 3`

`3`

## 说明

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.

`#include<cstdio>#include<cstring>#include<queue>using namespace std;const int MAXN=801,INF=5*1e8+10;inline char nc(){    static char buf[MAXN],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++;}inline int read(){    char c=nc();int x=0,f=1;    while(c<'0'||c>'9'){if(c=='-')f=-1;c=nc();}    while(c>='0'&&c<='9'){x=x*10+c-'0';c=nc();}    return x*f;}int S=0,T=301;struct node{    int u,v,flow,nxt;}edge[MAXN*20];int head[MAXN],cur[MAXN],num=0;inline void add_edge(int x,int y,int z){    edge[num].u=x;    edge[num].v=y;    edge[num].flow=z;    edge[num].nxt=head[x];    head[x]=num++;}inline void AddEdge(int x,int y,int z) {    add_edge(x,y,z);    add_edge(y,x,0);}int deep[MAXN];inline bool BFS(){    memset(deep,0,sizeof(deep));    deep[S]=1;    queue<int>q;    q.push(S);    while(q.size()!=0)    {        int p=q.front();        q.pop();        for(int i=head[p];i!=-1;i=edge[i].nxt)            if(!deep[edge[i].v]&&edge[i].flow)            {                deep[edge[i].v]=deep[p]+1;q.push(edge[i].v);                if(edge[i].v==T) return 1;            }    }    return deep[T];}int DFS(int now,int nowflow){    if(now==T||nowflow<=0)    return nowflow;    int totflow=0;    for(int &i=cur[now];i!=-1;i=edge[i].nxt)     {        if(deep[edge[i].v]==deep[now]+1&&edge[i].flow)        {            int canflow=DFS(edge[i].v,min(nowflow,edge[i].flow));            edge[i].flow-=canflow;edge[i^1].flow+=canflow;            totflow+=canflow;            nowflow-=canflow;            if(nowflow<=0) break;        }    }    return totflow;}int Dinic(){    int ans=0;    while(BFS())    {        memcpy(cur,head,sizeof(head));         ans+=DFS(S,INF);    }    return ans;}int N,F,D;int main(){    #ifdef WIN32    freopen("a.in","r",stdin);    #else    #endif    memset(head,-1,sizeof(head));    N=read();F=read();D=read();    for(int i=1;i<=N;i++)    {        int Fnum=read(),Dnum=read();        AddEdge(F+i,F+N+i,1);        for(int j=1;j<=Fnum;j++){int P=read();AddEdge(P,F+i,1);}        for(int j=1;j<=Dnum;j++){int P=read();AddEdge(F+N+i,2*N+F+P,1);};    }    for(int i=1;i<=F;i++) AddEdge(S,i,1);    for(int i=1;i<=D;i++) AddEdge(2*N+F+i,T,1);    printf("%d",Dinic());    return  0;}`

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