Dining
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 21578 Accepted: 9545
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.
Source
USACO 2007 Open Gold
看完题就知道是一道最大流，稍微思考一下就知道怎么建图了：这道题的难点就是每种饮料，每种食物和每头牛只能进行一次匹配。而这是许多最大流问题共同的套路，我们可以将每头牛拆成一个入口和出口，从入口向出口连一条容量为111的边，然后建一个源点sss和一个汇点ttt，从sss向每种食物（饮料）建一条容量为111的边，从每种饮料（食品）向ttt建一条容量为111的边，最后对于每种食品（饮料），向匹配的牛的入口建容量为111的边，对于每头牛的出口，向可以与之匹配的饮料（食品）连一条容量为111的边，最后从sss到ttt跑最大流即可
代码如下：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#define N 500
#define M 15000
using namespace std;
inline long long read(){
long long ans=0;
char ch=getchar();
while(!isdigit(ch))ch=getchar();
while(isdigit(ch))ans=(ans<<3)+(ans<<1)+ch-'0',ch=getchar();
return ans;
}
int n,f,d,s,t,dis[N],first[N],cnt=-1,ans=0;
struct Node{int v,next,c;}e[M];
inline void add(int u,int v,int c){
e[++cnt].v=v;
e[cnt].c=c;
e[cnt].next=first[u];
first[u]=cnt;
}
inline bool bfs(){
queue<int>q;
q.push(s);
memset(dis,-1,sizeof(dis));
dis[s]=0;
while(!q.empty()){
int x=q.front();
q.pop();
for(int i=first[x];i!=-1;i=e[i].next){
int v=e[i].v;
if(e[i].c<=0||dis[v]!=-1)continue;
dis[v]=dis[x]+1;
if(v==t)return true;
q.push(v);
}
}
return false;
}
inline int dfs(int x,int f){
if(!f||x==t)return f;
int flow=f;
for(int i=first[x];i!=-1;i=e[i].next){
int v=e[i].v;
if(flow&&e[i].c>0&&dis[v]==dis[x]+1){
int tmp=dfs(v,min(flow,e[i].c));
if(tmp==0)dis[v]=-1;
flow-=tmp;
e[i].c-=tmp;
e[i^1].c+=tmp;
}
}
return f-flow;
}
int main(){
memset(first,-1,sizeof(first));
n=read(),f=read(),d=read();
s=0,t=n+n+f+d+1;
for(int i=1;i<=f;++i)add(s,i,1),add(i,s,0);
for(int i=1;i<=d;++i)add(i+f+n+n,t,1),add(t,i+f+n+n,0);
for(int i=1;i<=n;++i){
int cntf=read(),cntd=read();
for(int j=1;j<=cntf;++j){
int v=read();
add(v,i+f,1),add(i+f,v,0);
}
for(int j=1;j<=cntd;++j){
int v=read();
add(i+f+n,v+f+n+n,1),add(v+f+n+n,i+f+n,0);
}
add(i+f,i+f+n,1),add(i+f+n,i+f,0);
}
while(bfs())ans+=dfs(s,0x3f3f3f3f);
printf("%d",ans);
return 0;
}