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| Time Limit: 2000MS | Memory Limit: 65536K | |
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
Output
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
分析
之前做过类似的题了，建图就是将奶牛拆成两个点，左边连食物，右边连饮料，食物和饮料分别连源点和汇点，跑最大流就行了。
#include<stdio.h> #include<string.h> #include<iostream> #include<queue> using namespace std; //**************************************************** //最大流模板Edmonds_Karp算法 //初始化：G[][],st,ed //****************************************************** const int MAXN = 500; const int INF = 0x3fffffff; int G[MAXN][MAXN];//存边的容量，没有边的初始化为0 int path[MAXN],flow[MAXN],st,ed; int n;//点的个数，编号0~n，n包括了源点和汇点 queue<int>q; int bfs() { int i,t; while(!q.empty()) q.pop();//清空队列 memset(path,-1,sizeof(path));//每次搜索前都把路径初始化成-1 path[st]=0; flow[st]=INF;//源点可以有无穷的流流进 q.push(st); while(!q.empty()){ t=q.front(); q.pop(); if(t==ed) break; for(i=0;i<=n;i++){ if(i!=st&&path[i]==-1&&G[t][i]){ flow[i]=flow[t]<G[t][i]?flow[t]:G[t][i]; q.push(i); path[i]=t; } } } if(path[ed]==-1) return -1;//即找不到汇点上去了。找不到增广路径了 return flow[ed]; } int Edmonds_Karp() { int max_flow=0; int step,now,pre; while((step=bfs())!=-1){ max_flow+=step; now=ed; while(now!=st){ pre=path[now]; G[pre][now]-=step; G[now][pre]+=step; now=pre; } } return max_flow; } int main() { int N,F,D; memset(G,0,sizeof(G)); scanf("%d%d%d",&N,&F,&D); n=2*N+F+D+1; st=0,ed=n; for(int i=1;i<=F;i++) G[st][i]=1; for(int i=1;i<=N;i++) G[F+i][N+F+i]=1; for(int i=1;i<=D;i++) G[F+2*N+i][ed]=1; for(int i=1;i<=N;i++){ int k1,k2,u; scanf("%d%d",&k1,&k2); for(int j=1;j<=k1;j++){ scanf("%d",&u); G[u][F+i]=1; } for(int j=1;j<=k2;j++){ scanf("%d",&u); G[F+N+i][2*N+F+u]=1; } } printf("%d\n",Edmonds_Karp()); return 0; }
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原文地址：http://www.cnblogs.com/wangdongkai/p/5618860.html
