【最大流模板题】POJ 3281 Dining

题目链接：http://poj.org/problem?id=3281

题目大意：给定N头牛、F个食物、D个饮料。每头牛有一定的喜好，只喜欢其中几个食物和饮料。每个食物和饮料只能给一头牛即一头牛只能得到一个食物和饮料。而且一头牛必须同时获得一个食物和一个饮料才能满足。问至多有多少头牛可以获得满足。

分析：刚开始想用二分匹配来写，但是有两个限制条件，同时获得食物和饮料才能满足。这样二分匹配显然不行。

       建立最大流模型求解：添加一个源点和一个汇点，按照源点->食物->牛->牛->饮料->汇点建立出模型，求解最大流即可。最大流建图是把食物和饮料放在两端。一头牛拆分成两个点，两点之间的容量为1（这样才能保证一头牛能获得一个食物和一个饮料）。把喜欢的食物和饮料跟牛建条边，容量为1。加个源点和汇点。源点与食物、饮料
和汇点的边容量都是1，表示每种食物和饮料只有一个。
       建图的时候要弄清楚点之间的关系，虽然看似简单，但感觉对刚接触的人来说还是不好想。

CODE：（Dinic算法）

#include<stdio.h>#include<algorithm>#include<string.h>#include<iostream>#include<vector>#include<queue>#define INF 0x3f3f3f3ftypedef long long LL;using namespace std;const int maxn=1008;struct edge{    int to,cap,rev;};vector<edge>G[maxn];//图的邻接表表示int iter[maxn];//顶点到源点的距离标号int level[maxn];//当前弧，在其之前的弧已经没有用了//向图中增加一条从from到to的容量为cap的边void addedge(int from,int to,int cap){    G[from].push_back((edge){to,cap,G[to].size()});    G[to].push_back((edge){from,0,G[from].size()-1});}//通过bfs计算从源点出发的距离标号void bfs(int s){    memset(level,-1,sizeof(level));    queue<int>que;    level[s]=0;    que.push(s);    while(!que.empty())    {        int v=que.front();        que.pop();        for(int i=0; i<G[v].size(); i++)        {            edge &e=G[v][i];            if(e.cap>0&&level[e.to]<0)            {                level[e.to]=level[v]+1;                que.push(e.to);            }        }    }}//通过dfs寻找增广路int dfs(int v,int t,int f){    if(v==t)        return f;    for(int &i=iter[v]; i<G[v].size(); i++)    {        edge &e=G[v][i];        if(e.cap>0&&level[v]<level[e.to])        {            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;}//求解从s到t的最大流int max_flow(int s,int t){    int flow=0;    for(;;)    {        bfs(s);        if(level[t]<0)            return flow;        memset(iter,0,sizeof(iter));        int f;        while((f=dfs(s,t,INF))>0)        {            flow+=f;            //cout<<f<<" ";        }    }}int main(){    int n,f,d,s,t;    while(~scanf("%d%d%d",&n,&f,&d))    {        s=0,t=d+f+2*n+1;        for(int i=1; i<=t; i++)            G[i].clear();        for(int i=1; i<=f; i++)            addedge(s,i,1);//源点跟食物之间        int fi,di,x;        for(int i=1; i<=n; i++)        {            addedge(f+i,f+i+n,1);//一头牛之间分两个点建立容量为1的边            scanf("%d%d",&fi,&di);            for(int j=1; j<=fi; j++)            {                scanf("%d",&x);                addedge(x,f+i,1);//食物跟牛之间            }            for(int j=1; j<=di; j++)            {                scanf("%d",&x);                addedge(f+i+n,f+2*n+x,1);//饮料跟牛之间            }        }        for(int i=1; i<=d; i++)            addedge(f+2*n+i,t,1);//牛与汇点之间        int ans=max_flow(s,t);        printf("%d\n",ans);    }    return 0;}

题目描述：

Dining

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 18155 Accepted: 8092

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 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 32 2 1 2 3 12 2 2 3 1 22 2 1 3 1 22 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.

Source

USACO 2007 Open Gold