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题意:
有F种食物和D种饮料,每头牛有各自喜欢的食物和饮料,而且每种食物或者饮料只能给一头牛。
求最多能有多少头牛能同时得到它喜欢的食物或者饮料。
分析:
把每个牛拆点,中间连一条容量为1的边,保证一头牛不会被多个食物或者饮料分配。
然后把饮料和牛连边,食物和另外一边的牛连边,最后增加一个源点和汇点跑最大流。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #include <queue> 7 using namespace std; 8 9 const int maxn = 100 + 10; 10 const int maxnode = 400 + 10; 11 const int INF = 0x3f3f3f3f; 12 13 int N, F, D; 14 15 int n, s, t; 16 17 struct Edge 18 { 19 int from, to, cap, flow; 20 Edge(int u, int v, int c, int f):from(u), to(v), cap(c), flow(f) {} 21 }; 22 23 vector<Edge> edges; 24 vector<int> G[maxnode]; 25 26 void init() 27 { 28 edges.clear(); 29 for(int i = 0; i < n; i++) G[i].clear(); 30 } 31 32 void AddEdge(int u, int v, int c) 33 { 34 edges.push_back(Edge(u, v, c, 0)); 35 edges.push_back(Edge(v, u, 0, 0)); 36 int m = edges.size(); 37 G[u].push_back(m - 2); 38 G[v].push_back(m - 1); 39 } 40 41 bool vis[maxnode]; 42 int d[maxnode]; 43 int cur[maxnode]; 44 45 bool BFS() 46 { 47 d[s] = 0; 48 queue<int> Q; 49 Q.push(s); 50 memset(vis, false, sizeof(vis)); 51 vis[s] = true; 52 53 while(!Q.empty()) 54 { 55 int u = Q.front(); Q.pop(); 56 for(int i = 0; i < G[u].size(); i++) 57 { 58 Edge& e = edges[G[u][i]]; 59 int v = e.to; 60 if(!vis[v] && e.cap > e.flow) 61 { 62 vis[v] = true; 63 d[v] = d[u] + 1; 64 Q.push(v); 65 } 66 } 67 } 68 69 return vis[t]; 70 } 71 72 int DFS(int u, int a) 73 { 74 if(u == t || a == 0) return a; 75 int flow = 0, f; 76 for(int& i = cur[u]; i < G[u].size(); i++) 77 { 78 Edge& e = edges[G[u][i]]; 79 int v = e.to; 80 if(d[v] == d[u] + 1 && (f = DFS(v, min(a, e.cap - e.flow))) > 0) 81 { 82 flow += f; 83 e.flow += f; 84 a -= f; 85 edges[G[u][i]^1].flow -= f; 86 if(a == 0) break; 87 } 88 } 89 return flow; 90 } 91 92 int Maxflow() 93 { 94 int flow = 0; 95 while(BFS()) 96 { 97 memset(cur, 0, sizeof(cur)); 98 flow += DFS(s, INF); 99 } 100 return flow; 101 } 102 103 int main() 104 { 105 while(scanf("%d%d%d", &N, &F, &D) == 3) 106 { 107 n = N * 2 + D + F + 2; 108 s = 0, t = n - 1; 109 init(); 110 111 //build graph 112 for(int i = 1; i <= F; i++) AddEdge(s, N*2+i, 1); 113 for(int i = 1; i <= D; i++) AddEdge(N*2+F+i, t, 1); 114 for(int i = 1; i <= N; i++) AddEdge(i, i + N, 1); 115 116 for(int i = 1; i <= N; i++) 117 { 118 int f, d, x; scanf("%d%d", &f, &d); 119 while(f--) 120 { 121 scanf("%d", &x); 122 AddEdge(N*2+x, i, 1); 123 } 124 while(d--) 125 { 126 scanf("%d", &x); 127 AddEdge(N + i, N*2+F+x, 1); 128 } 129 } 130 131 printf("%d\n", Maxflow()); 132 } 133 134 return 0; 135 }
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原文地址:http://www.cnblogs.com/AOQNRMGYXLMV/p/4793999.html