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
[Submit][Status][Discuss]

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

[Submit][Status][Discuss]

 

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……

 

  1 #include <cstdio>
  2 
  3 inline int nextChar(void) {
  4     const int siz = 1024;
  5     
  6     static char buf[siz];
  7     static char *hd = buf + siz;
  8     static char *tl = buf + siz;
  9     
 10     if (hd == tl)
 11         fread(hd = buf, 1, siz, stdin);
 12         
 13     return *hd++;
 14 }
 15  
 16 inline int nextInt(void) {
 17     register int ret = 0;
 18     register int neg = false;
 19     register int bit = nextChar();
 20     
 21     for (; bit < 48; bit = nextChar())
 22         if (bit == '-')neg ^= true;
 23         
 24     for (; bit > 47; bit = nextChar())
 25         ret = ret * 10 + bit - 48;
 26         
 27     return neg ? -ret : ret;
 28 }
 29 
 30 inline int min(int a, int b)
 31 {
 32     return a < b ? a : b;
 33 }
 34 
 35 const int siz = 500005;
 36 const int inf = 1000000007;
 37 
 38 int tot;
 39 int s, t;
 40 int hd[siz];
 41 int to[siz];
 42 int fl[siz];
 43 int nt[siz];
 44 
 45 inline void add(int u, int v, int f)
 46 {
 47     nt[tot] = hd[u]; to[tot] = v; fl[tot] = f; hd[u] = tot++;
 48     nt[tot] = hd[v]; to[tot] = u; fl[tot] = 0; hd[v] = tot++;
 49 }
 50 
 51 int dep[siz];
 52 
 53 inline bool bfs(void)
 54 {
 55     static int que[siz], head, tail;
 56     
 57     for (int i = s; i <= t; ++i)dep[i] = 0;
 58     
 59     dep[que[head = 0] = s] = tail = 1;
 60     
 61     while (head != tail)
 62     {
 63         int u = que[head++], v;
 64         
 65         for (int i = hd[u]; ~i; i = nt[i])
 66             if (!dep[v = to[i]] && fl[i])
 67                 dep[que[tail++] = v] = dep[u] + 1;
 68     }
 69     
 70     return dep[t];
 71 }
 72 
 73 int cur[siz];
 74 
 75 int dfs(int u, int f)
 76 {
 77     if (u == t || !f)
 78         return f;
 79         
 80     int used = 0, flow, v;
 81     
 82     for (int i = cur[u]; ~i; i = nt[i])
 83         if (dep[v = to[i]] == dep[u] + 1 && fl[i])
 84         {
 85             flow = dfs(v, min(fl[i], f - used));
 86             
 87             used += flow;
 88             fl[i] -= flow;
 89             fl[i^1] += flow;
 90             
 91             if (used == f)
 92                 return f;
 93             
 94             if (fl[i])
 95                 cur[u] = i;
 96         }
 97         
 98     if (!used)
 99         dep[u] = 0;
100     
101     return used;
102 }
103 
104 inline int maxFlow(void)
105 {
106     int maxFlow = 0, newFlow;
107     
108     while (bfs())
109     {
110         for (int i = s; i <= t; ++i)
111             cur[i] = hd[i];
112         
113         while (newFlow = dfs(s, inf))
114             maxFlow += newFlow;
115     }
116     
117     return maxFlow;
118 }
119 
120 int N, F, D;
121 
122 inline int cow(int x, int y)
123 {
124     return F + D + y * N + x;
125 }
126 
127 inline int food(int x)
128 {
129     return x;
130 }
131 
132 inline int drink(int x)
133 {
134     return x + F;
135 }
136 
137 signed main(void)
138 {
139     N = nextInt();
140     F = nextInt();
141     D = nextInt();
142     
143     s = 0, t = N*2 + F + D + 1;
144     
145     for (int i = s; i <= t; ++i)
146         hd[i] = -1;
147         
148     for (int i = 1; i <= F; ++i)
149         add(s, food(i), 1);
150         
151     for (int i = 1; i <= D; ++i)
152         add(drink(i), t, 1);
153         
154     for (int i = 1; i <= N; ++i)
155     {
156         int f = nextInt();
157         int d = nextInt();
158         
159         add(cow(i, 0), cow(i, 1), 1);
160         
161         for (int j = 1; j <= f; ++j)
162             add(food(nextInt()), cow(i, 0), 1);
163         
164         for (int j = 1; j <= d; ++j)
165             add(cow(i, 1), drink(nextInt()), 1);
166     }
167     
168     printf("%d\n", maxFlow());
169 }

 

@Author: YouSiki

posted @ 2017-01-05 20:10  YouSiki  阅读(289)  评论(0)  编辑  收藏  举报