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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 10768 | Accepted: 4938 |
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
Hint
Source
1 #include<cstring> 2 #include<cstdio> 3 #include<algorithm> 4 #include<vector> 5 #include<queue> 6 7 using namespace std; 8 9 const int maxn=405; 10 const int inf=0x3f3f3f3f; 11 12 struct Edge 13 { 14 int to,cap,rev; 15 }; 16 17 vector<Edge>edge[maxn]; 18 int level[maxn]; 19 int id[maxn]; 20 int s=0; 21 int t; 22 23 void addedge(int from,int to) 24 { 25 edge[from].push_back((Edge){to,1,edge[to].size()}); 26 edge[to].push_back((Edge){from,0,edge[from].size()-1}); 27 } 28 29 void bfs() 30 { 31 memset(level,-1,sizeof(level)); 32 33 queue<int>que; 34 while(!que.empty()) 35 que.pop(); 36 37 level[s]=0; 38 que.push(s); 39 40 while(!que.empty()) 41 { 42 int u=que.front(); 43 que.pop(); 44 for(int i=0;i<edge[u].size();i++) 45 { 46 Edge &e=edge[u][i]; 47 if(e.cap>0&&level[e.to]<0) 48 { 49 level[e.to]=level[u]+1; 50 que.push(e.to); 51 } 52 } 53 } 54 } 55 56 int dfs(int u,int f) 57 { 58 if(u==t) 59 return f; 60 for(int &i=id[u];i<edge[u].size();i++) 61 { 62 Edge &e=edge[u][i]; 63 if(e.cap>0&&level[e.to]>level[u]) 64 { 65 int d=dfs(e.to,min(e.cap,f)); 66 if(d>0) 67 { 68 e.cap-=d; 69 edge[e.to][e.rev].cap+=d; 70 return d; 71 } 72 } 73 } 74 return 0; 75 } 76 77 int solve() 78 { 79 int res=0; 80 while(true) 81 { 82 bfs(); 83 if(level[t]<0) 84 return res; 85 memset(id,0,sizeof(id)); 86 int flow; 87 while(flow=dfs(s,inf)) 88 { 89 res+=flow; 90 } 91 } 92 } 93 94 int main() 95 { 96 int N,F,D; 97 while(~scanf("%d%d%d",&N,&F,&D)) 98 { 99 for(int i=0;i<maxn;i++) 100 edge[i].clear(); 101 102 t=F+2*N+D+1; 103 for(int i=1;i<=F;i++) 104 addedge(s,i); 105 for(int i=1;i<=D;i++) 106 addedge(F+2*N+i,t); 107 for(int i=1;i<=N;i++) 108 addedge(i+F,i+F+N); 109 for(int i=1;i<=N;i++) 110 { 111 int a,b; 112 scanf("%d%d",&a,&b); 113 for(int j=1;j<=a;j++) 114 { 115 int u; 116 scanf("%d",&u); 117 addedge(u,i+F); 118 } 119 for(int j=1;j<=b;j++) 120 { 121 int u; 122 scanf("%d",&u); 123 addedge(i+F+N,u+F+2*N); 124 } 125 } 126 127 printf("%d\n",solve()); 128 } 129 return 0; 130 }
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原文地址：http://www.cnblogs.com/-maybe/p/4576009.html
