Dining Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 17557 Accepted: 7822
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
#include<iostream> #include<cstring> #include<cstdio> #include<queue> using namespace std; #define maxm 500000 #define maxn 100000 int S,T,F,D,N; int head[maxn],dep[maxn],tot=1; struct Edge{ int to,value,next; }e[maxm]; void Add_Edge(int u,int v,int w){ e[++tot].to=v;e[tot].value=w; e[tot].next=head[u];head[u]=tot; } queue<int> q; bool BFS(){ memset(dep,-1,sizeof dep ); dep[S]=0;q.push(S); while(!q.empty()){ int u=q.front();q.pop(); for(int i=head[u];i;i=e[i].next){ int v=e[i].to; if(dep[v]==-1&&e[i].value){ dep[v]=dep[u]+1; q.push(v); } } } if(dep[T]!=-1) return true; else return false; } int DFS(int u,int flow){ int ret=0; if(u==T) return flow; for(int i=head[u];i;i=e[i].next){ int v=e[i].to; if(dep[v]==dep[u]+1&&e[i].value){ int x=DFS(v,min(e[i].value,flow)); ret+=x;flow-=x; e[i].value-=x;e[i^1].value+=x; } } return ret; } void Dinic(){ int ans=0; while(BFS()) ans+=DFS(S,1e9); printf("%d\n",ans); return; } int main(){ scanf("%d%d%d",&N,&F,&D); T=N+N+F+D+20;S=0;int x; for(int i=1;i<=F;i++) Add_Edge(S,i,1),Add_Edge(i,S,0); for(int i=1,sumf,sumd;i<=N;i++){ scanf("%d%d",&sumf,&sumd); for(int j=1;j<=sumf;j++){ scanf("%d",&x);// Add_Edge(x,i+F,1); Add_Edge(i+F,x,0); } for(int x,j=1;j<=sumd;j++){ scanf("%d",&x); Add_Edge(i+F+N,N+N+F+x,1); Add_Edge(N+N+F+x,i+F+N,0); } } for(int i=1;i<=N;i++) Add_Edge(i+F,i+F+N,1),Add_Edge(i+F+N,i+F,0); for(int i=1;i<=D;i++) Add_Edge(i+F+N+N,T,1),Add_Edge(T,i+F+N+N,0); Dinic(); return 0; }源点连边向食物，食物连边向牛，牛连边向饮料，饮料连边向汇点，跑Dinic，注意中间牛拆一下点不然可能会一头牛分配多个食物
