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
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