BZOJ 1711: [Usaco2007 Open]Dining吃饭

1711: [Usaco2007 Open]Dining吃饭

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 902  Solved: 476
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Description

农夫JOHN为牛们做了很好的食品,但是牛吃饭很挑食. 每一头牛只喜欢吃一些食品和饮料而别的一概不吃.虽然他不一定能把所有牛喂饱,他还是想让尽可能多的牛吃到他们喜欢的食品和饮料. 农夫JOHN做了F (1 <= F <= 100) 种食品并准备了D (1 <= D <= 100) 种饮料. 他的N (1 <= N <= 100)头牛都以决定了是否愿意吃某种食物和喝某种饮料. 农夫JOHN想给每一头牛一种食品和一种饮料,使得尽可能多的牛得到喜欢的食物和饮料. 每一件食物和饮料只能由一头牛来用. 例如如果食物2被一头牛吃掉了,没有别的牛能吃食物2.

Input

* 第一行: 三个数: N, F, 和 D

* 第2..N+1行: 每一行由两个数开始F_i 和 D_i, 分别是第i 头牛可以吃的食品数和可以喝的饮料数.下F_i个整数是第i头牛可以吃的食品号,再下面的D_i个整数是第i头牛可以喝的饮料号码.

Output

* 第一行: 一个整数,最多可以喂饱的牛数.

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

输入解释:

牛 1: 食品从 {1,2}, 饮料从 {1,2} 中选
牛 2: 食品从 {2,3}, 饮料从 {1,2} 中选
牛 3: 食品从 {1,3}, 饮料从 {1,2} 中选
牛 4: 食品从 {1,3}, 饮料从 {3} 中选

Sample Output

3
输出解释:

一个方案是:
Cow 1: 不吃
Cow 2: 食品 #2, 饮料 #2
Cow 3: 食品 #1, 饮料 #1
Cow 4: 食品 #3, 饮料 #3
用鸽笼定理可以推出没有更好的解 (一共只有3总食品和饮料).当然,别的数据会更难.

HINT

 

Source

Gold

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网络流，拆点建图，每一条流都对应着满足一头牛的方案。和今天考试T1贼像……

 #include <cstdio>

 inline int nextChar(void) {
     const int siz = ;

     static char buf[siz];
     static char *hd = buf + siz;
     static char *tl = buf + siz;

     if (hd == tl)
         fread(hd = buf, , siz, stdin);

     return *hd++;
 }

 inline int nextInt(void) {
     register int ret = ;
     register int neg = false;
     register int bit = nextChar();

     for (; bit < ; bit = nextChar())
         if (bit == '-')neg ^= true;

     for (; bit > ; bit = nextChar())
         ret = ret *  + bit - ;

     return neg ? -ret : ret;
 }

 inline int min(int a, int b)
 {
     return a < b ? a : b;
 }

 const int siz = ;
 const int inf = ;

 int tot;
 int s, t;
 int hd[siz];
 int to[siz];
 int fl[siz];
 int nt[siz];

 inline void add(int u, int v, int f)
 {
     nt[tot] = hd[u]; to[tot] = v; fl[tot] = f; hd[u] = tot++;
     nt[tot] = hd[v]; to[tot] = u; fl[tot] = ; hd[v] = tot++;
 }

 int dep[siz];

 inline bool bfs(void)
 {
     static int que[siz], head, tail;

     for (int i = s; i <= t; ++i)dep[i] = ;

     dep[que[head = ] = s] = tail = ;

     while (head != tail)
     {
         int u = que[head++], v;

         for (int i = hd[u]; ~i; i = nt[i])
             if (!dep[v = to[i]] && fl[i])
                 dep[que[tail++] = v] = dep[u] + ;
     }

     return dep[t];
 }

 int cur[siz];

 int dfs(int u, int f)
 {
     if (u == t || !f)
         return f;

     int used = , flow, v;

     for (int i = cur[u]; ~i; i = nt[i])
         if (dep[v = to[i]] == dep[u] +  && fl[i])
         {
             flow = dfs(v, min(fl[i], f - used));

             used += flow;
             fl[i] -= flow;
             fl[i^] += flow;

             if (used == f)
                 return f;

             if (fl[i])
                 cur[u] = i;
         }

     if (!used)
         dep[u] = ;

     return used;
 }

 inline int maxFlow(void)
 {
     int maxFlow = , newFlow;

     while (bfs())
     {
         for (int i = s; i <= t; ++i)
             cur[i] = hd[i];

         while (newFlow = dfs(s, inf))
             maxFlow += newFlow;
     }

     return maxFlow;
 }

 int N, F, D;

 inline int cow(int x, int y)
 {
     return F + D + y * N + x;
 }

 inline int food(int x)
 {
     return x;
 }

 inline int drink(int x)
 {
     return x + F;
 }

 signed main(void)
 {
     N = nextInt();
     F = nextInt();
     D = nextInt();

     s = , t = N* + F + D + ;

     for (int i = s; i <= t; ++i)
         hd[i] = -;

     for (int i = ; i <= F; ++i)
         add(s, food(i), );

     for (int i = ; i <= D; ++i)
         add(drink(i), t, );

     for (int i = ; i <= N; ++i)
     {
         int f = nextInt();
         int d = nextInt();

         add(cow(i, ), cow(i, ), );

         for (int j = ; j <= f; ++j)
             add(food(nextInt()), cow(i, ), );

         for (int j = ; j <= d; ++j)
             add(cow(i, ), drink(nextInt()), );
     }

     printf("%d\n", maxFlow());
 }

@Author: YouSiki