Dining POJ - 3281

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 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. 

 

题解：建图（神来之笔啊！），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 s 和一个汇点 t 。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。边的方向为 s-->食物-->牛-->牛-->饮料-->t ，容量全为1.

                                                                                               ----------------------------摘自《挑战程序设计竞赛》

真想甩几个大拇指，果然。。。不是现在我能想到的·~~`

#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn=1005;
const int INF=0x3f3f3f3f;

struct edge{
    int to,w;
    int nxt,rev;
}e[maxn*10];

int N,F,D,tot;
int head[maxn],vis[maxn];

void Init(){
    tot=0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int w){
    e[tot].to=v,e[tot].w=w,e[tot].nxt=head[u],e[tot].rev=tot+1,head[u]=tot++;
    e[tot].to=u,e[tot].w=0,e[tot].nxt=head[v],e[tot].rev=tot-1,head[v]=tot++;
}

int DFS(int s,int t,int f){
    if(s==t) return f;
    vis[s]=true;
    for(int i=head[s];i!=-1;i=e[i].nxt){
        int v=e[i].to;
        if(!vis[v]&&e[i].w>0){
            int d=DFS(v,t,min(e[i].w,f));               //改成 min(f,e[i].w)会多30多毫秒，？？？？
            if(d<=0) continue;
            e[i].w-=d;
            e[e[i].rev].w+=d;
            return d;
        }
    }
    return 0;
}

int max_Flow(int s,int t){
    int ans=0;
    while(true){
        memset(vis,false,sizeof(vis));
        int temp=DFS(s,t,INF);
        if(temp==0) return ans;
        ans=ans+temp;
    }
}

void solve(){
    int s=0,t=F+2*N+D+1;
    for(int i=1;i<=F;i++) addedge(s,i,1);
    for(int i=1;i<=D;i++) addedge(F+2*N+i,t,1);
    for(int i=1;i<=N;i++) addedge(F+i,F+i+N,1);
    for(int i=1;i<=N;i++) {
        int p,q,x;
        cin>>p>>q;
        for(int j=1;j<=p;j++){ cin>>x; addedge(x,F+i,1); }
        for(int j=1;j<=q;j++){ cin>>x; addedge(F+N+i,F+2*N+x,1);} 
    }
    printf("%d\n",max_Flow(s,t));
}

int main()
{   while(cin>>N>>F>>D){ Init(); solve(); }
    return 0;
}

/*这种写法会超时，让我很无语
void Read(){
    cin>>N>>F>>D;
    int p,q,x;
    for(int i=0;i<N;i++){
        cin>>p>>q;
        for(int j=0;j<p;j++){ cin>>x; likeF[i][x-1]=true; }
        for(int j=0;j<q;j++){ cin>>x; likeD[i][x-1]=true; } 
    }
}

void solve(){
    int s=N*2+F+D;
    int t=s+1;
    for(int i=0;i<F;i++) addedge(s,N*2+i,1);
    for(int i=0;i<D;i++) addedge(N*2+F+i,t,1);
    for(int i=0;i<N;i++){
        addedge(i,N+i,1);
        for(int j=0;j<F;j++) if(likeF[i][j]) addedge(N*2+j,i,1);
        for(int j=0;j<D;j++) if(likeD[i][j]) addedge(N+i,N*2+F+j,1); 
    }
    printf("%d\n",max_Flow(s,t));
}
*/