洛谷P2891 [USACO07OPEN]吃飯Dining
題目描述
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).
有F種食物和D種飲料，每種食物或飲料只能供一頭牛享用，且每頭牛只享用一種食物和一種飲料。現在有n頭牛，每頭牛都有自己喜歡的食物種類列表和飲料種類列表，問最多能使幾頭牛同時享用到自己喜歡的食物和飲料。（1 <= f <= 100, 1 <= d <= 100, 1 <= n <= 100）
輸入輸出格式
輸入格式：
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.
輸出格式：
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
輸入輸出樣例
輸入樣例#1： 復制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輸出樣例#1： 復制
3
說明
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.
最大流問題。
一個比較容易想到的思路就是
但是這樣可能出現左邊有兩個流量經過中間同一個點的情況
所以我們把中間的點拆開
最終的圖應該是這樣的
#include<cstdio> #include<cstring> #include<queue> using namespace std; const int MAXN=801,INF=5*1e8+10; inline char nc() { static char buf[MAXN],*p1=buf,*p2=buf; return p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++; } inline int read() { char c=nc();int x=0,f=1; while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=nc();} while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=nc();} return x*f; } int S=0,T=301; struct node { int u,v,flow,nxt; }edge[MAXN*20]; int head[MAXN],cur[MAXN],num=0; inline void add_edge(int x,int y,int z) { edge[num].u=x; edge[num].v=y; edge[num].flow=z; edge[num].nxt=head[x]; head[x]=num++; } inline void AddEdge(int x,int y,int z) { add_edge(x,y,z); add_edge(y,x,0); }int deep[MAXN]; inline bool BFS() { memset(deep,0,sizeof(deep)); deep[S]=1; queue<int>q; q.push(S); while(q.size()!=0) { int p=q.front(); q.pop(); for(int i=head[p];i!=-1;i=edge[i].nxt) if(!deep[edge[i].v]&&edge[i].flow) { deep[edge[i].v]=deep[p]+1;q.push(edge[i].v); if(edge[i].v==T) return 1; } } return deep[T]; } int DFS(int now,int nowflow) { if(now==T||nowflow<=0) return nowflow; int totflow=0; for(int &i=cur[now];i!=-1;i=edge[i].nxt) { if(deep[edge[i].v]==deep[now]+1&&edge[i].flow) { int canflow=DFS(edge[i].v,min(nowflow,edge[i].flow)); edge[i].flow-=canflow;edge[i^1].flow+=canflow; totflow+=canflow; nowflow-=canflow; if(nowflow<=0) break; } } return totflow; } int Dinic() { int ans=0; while(BFS()) { memcpy(cur,head,sizeof(head)); ans+=DFS(S,INF); } return ans; } int N,F,D; int main() { #ifdef WIN32 freopen("a.in","r",stdin); #else #endif memset(head,-1,sizeof(head)); N=read();F=read();D=read(); for(int i=1;i<=N;i++) { int Fnum=read(),Dnum=read(); AddEdge(F+i,F+N+i,1); for(int j=1;j<=Fnum;j++){int P=read();AddEdge(P,F+i,1);} for(int j=1;j<=Dnum;j++){int P=read();AddEdge(F+N+i,2*N+F+P,1);}; } for(int i=1;i<=F;i++) AddEdge(S,i,1); for(int i=1;i<=D;i++) AddEdge(2*N+F+i,T,1); printf("%d",Dinic()); return 0; }
洛谷P2891 [USACO07OPEN]吃飯Dining