POJ 3281 Dining(网络流-拆点)

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 Fiintegers 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.
题解：题目给你N头牛，每头牛都有喜欢的饮料喝喜欢的食物，而每一种食物和饮料都只能给一头牛，让你求，最多可以分配多少头牛食物饮料；
由于每种饮料喝每种食物只能给一种牛，因此我们可以将牛这个节点拆分为一条容量为1的边，然后让牛喝饮料喝食物分别建立一条容量为1的边，这样就转化为
最大流的问题；
参考代码为：
  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<cmath>
  5 #include<cstdlib>
  6 #include<algorithm>
  7 #include<queue>
  8 #include<deque>
  9 #include<stack>
 10 #include<set>
 11 #include<vector>
 12 #include<map>
 13 using namespace std;
 14 typedef long long LL;
 15 typedef pair<int,int> PII;
 16 #define PI acos(-1)
 17 #define EPS 1e-8
 18 const int INF=0x3f3f3f3f;
 19 const int maxn = 520;
 20 int n,m,k,s,t,x,y,z;
 21 struct Edge { 
 22     int from, to, cap, flow;
 23 };
 24 vector<Edge> edges;
 25 vector<int> G[maxn];
 26 bool vis[maxn];
 27 int d[maxn], cur[maxn];
 28 
 29 void Init()
 30 {
 31     memset(d,0,sizeof d);
 32     for(int i=0;i<=n;i++) G[i].clear();
 33 }
 34 
 35 void addedge(int from, int to, int cap) 
 36 {
 37     edges.push_back((Edge){from, to, cap, 0});
 38     edges.push_back((Edge){to, from, 0, 0});
 39     int m = edges.size();
 40     G[from].push_back(m-2); G[to].push_back(m-1);
 41 }
 42 
 43 bool bfs() 
 44 {
 45     memset(vis,0,sizeof vis);
 46     queue<int> q;
 47     q.push(s);
 48     d[s] = 0; vis[s] = 1;
 49     while (!q.empty()) 
 50     {
 51         int x = q.front(); q.pop();
 52         for(int i = 0; i < G[x].size(); ++i) 
 53         {
 54             Edge &e = edges[G[x][i]];
 55             if (!vis[e.to] && e.cap > e.flow) 
 56             {
 57                 vis[e.to] = 1;
 58                 d[e.to] = d[x] + 1;
 59                 q.push(e.to);
 60             }
 61         }
 62     }
 63     return vis[t];
 64 }
 65 
 66 int dfs(int x,int a) 
 67 {
 68     if(x == t || a == 0) return a;
 69     int flow = 0, f;
 70     for(int &i = cur[x]; i < G[x].size(); ++i) 
 71     {
 72         Edge &e = edges[G[x][i]];
 73         if (d[e.to] == d[x] + 1 && (f=dfs(e.to, min(a, e.cap-e.flow))) > 0) 
 74         {
 75             e.flow += f;
 76             edges[G[x][i]^1].flow -= f;
 77             flow += f; a -= f;
 78             if (a == 0) break;
 79         }
 80     }
 81     return flow;
 82 }
 83 
 84 int Maxflow(int s, int t) 
 85 {
 86     int flow = 0;
 87     while (bfs()) 
 88     {
 89         memset(cur,0,sizeof cur);
 90         flow += dfs(s, INF);
 91     }
 92     return flow;
 93 }
 94 
 95 int main()
 96 {
 97     while(~scanf("%d%d%d",&n,&m,&k))
 98     {
 99         Init();
100         s=0,t=2*n+m+k+1;
101         for(int i=1;i<=m;i++) addedge(s,i,1);
102         for(int i=1;i<=n;i++) addedge(m+i,n+m+i,1);
103         for(int i=2*n+m+1;i<=2*n+m+k;i++) addedge(i,t,1);
104         for(int i=1;i<=n;i++)
105         {
106             scanf("%d%d",&x,&y);
107             for(int j=1;j<=x;j++) scanf("%d",&z),addedge(z,m+i,1);
108             for(int j=1;j<=y;j++) scanf("%d",&z),addedge(m+n+i,z+m+2*n,1);
109         }
110         printf("%d\n",Maxflow(s,t));
111     }
112     return 0;
113 }
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posted @ 2018-08-23 20:36  StarHai  阅读(173)  评论(0)  编辑  收藏  举报