Dining POJ3281 最大流

来自《挑战程序设计竞赛》

1.题目原文

http://poj.org/problem?id=3281

Language:Default

Dining

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 15462 Accepted: 7012

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

2.解题思路

如果只是分配食物的话，那么直接用二分图最大匹配就能够解决了。但是遇到这种需要同时给一头牛分配所喜欢的食物和饮料，就不能很好地处理。我们可以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，还有一个源点s，一个汇点t

②在两个匹配相同的牛之间连一条边，在s和所有食物之间连一条边，在所有饮料和t之间连一条边。

③边的方向为s→ 食物→ 牛→ 牛→ 饮料→ t。

这个图中的每一条路径都对应一个牛的食物和饮料的匹配方案。问题转化为最大流问题。

3.AC代码

#include<iostream>#include<cstdio>#include<vector>#include<cstring>#include<algorithm>#include<utility>#include<queue>using namespace std;#define INF 0x7fffffff#define MAX_N 105#define MAX_F 105#define MAX_D 105//最大流算法Dinic模板#define MAX_V 405//用于表示边的结构体(终点、容量、反向边)struct edge{    int to,cap,rev;};vector<edge> G[MAX_V];//图的邻接表表示int level[MAX_V];//顶点到源点的距离标号int iter[MAX_V];//当前弧，在其之前的边已经没有用了//向图中增加一条从from到to的容量为cap的边void add_edge(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;        }    }}//模板int N,F,D;bool likeF[MAX_N][MAX_F];//食物的喜好bool likeD[MAX_N][MAX_D];//饮料的喜好void solve(){    //0-N-1:食物一侧的牛    //N-2N-1:饮料一侧的牛    //2N-2N+F-1:食物    //2N+F-2N+F+D-1:饮料    int s=2*N+F+D,t=s+1;    //在s和食物之间连边    for(int i=0;i<F;i++){        add_edge(s,2*N+i,1);    }    //在饮料和t之间连边    for(int i=0;i<D;i++){        add_edge(2*N+F+i,t,1);    }    for(int i=0;i<N;i++){        //在食物一侧的牛和饮料一侧的牛之间连边        add_edge(i,N+i,1);        //在牛和所喜欢的食物或者饮料之间连边        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);            }        }    }    printf("%d\n",max_flow(s,t));}int main(){    memset(likeF,0,sizeof(likeF));    memset(likeD,0,sizeof(likeD));    scanf("%d%d%d",&N,&F,&D);    for(int i=0;i<N;i++){        int n,m,f,d;        scanf("%d%d",&n,&m);        for(int j=0;j<n;j++){            scanf("%d",&f);            likeF[i][f-1]=true;        }        for(int j=0;j<m;j++){            scanf("%d",&d);            likeD[i][d-1]=true;        }    }    solve();    return 0;}