[USACO5.3]校园网Network of Schools

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[Time Gate] 

https://www.luogu.org/problemnew/show/P2746

【解题思路】

Tarjan缩点
第一问就过于水了,缩完点后,输出入度为0的点的个数即可。
第二问略微麻烦,考虑出度为零的点是一定不合法的,想让它合法,最有效率的一种方法就是让他连到一个入度为零的点上,而且如果有选择的余地的话,尽量让他连到一个走不到它的入度为0的点上,如果都能走到它那随便连也没问题。 当然入度为零的点多于出度为零的点的话也是不合法的,那就找一个出度为零(可以重复)的点连过去就好了。
设入度为零的点有cnt1个,出度为零的有cnt2个。
则答案为:max(cnt1,cnt2)

【code】

 

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <cctype>
 6 #include <cmath>
 7 
 8 using namespace std;
 9 
10 const int N = 1020;
11 
12 int n, ec = -1, a, b, id = 0, top, cur, ans = 0, color = 0, cnt;
13 int siz[N], dfn[N], f[N], low[N], stk[N], in[N], vis[N], col[N], out[N];
14 
15 struct Edge
16 
17     int u, v, nxt;
18 e[N << 2];
19 
20 inline void Addedge(int a, int b)
21 
22     ++ec;
23     e[ec].u = a;
24     e[ec].v = b;
25     e[ec].nxt = f[a];
26     f[a] = ec;
27 
28 
29 inline void read(int &x)
30 
31     int k = 1; x = 0;
32     char c = getchar();
33     while (!isdigit(c)) 
34         if (c == -) c = getchar(), k = -1;
35         else c = getchar();
36     while (isdigit(c))
37         x = (x << 1) + (x << 3) + (c ^ 48),
38         c = getchar();
39     x *= k;
40 
41 
42 void Tarjan(int u)
43 
44     dfn[u] = low[u] = ++id;
45     stk[++top] = u, vis[u] = 1;
46     for (int i = f[u]; i != -1; i = e[i].nxt)
47         if (!dfn[e[i].v])
48             Tarjan(e[i].v), low[u] = min(low[u], low[e[i].v]);
49         else if (vis[e[i].v])
50             low[u] = min(low[u], dfn[e[i].v]);
51     if (dfn[u] == low[u])
52     
53         ++color;
54         do
55         
56             cur = stk[top--];
57             vis[cur] = 0;
58             ++siz[color];
59             col[cur] = color;
60         while (cur != u);
61     
62 
63 
64 signed main()
65 
66     read(n);
67     memset(f, -1, sizeof(f));
68     for (int i = 1; i <= n; ++i)
69         while (true)
70         
71             read(b);
72             if (!b) break;
73             Addedge(i, b);
74         
75     for (int i = 1; i <= n; ++i)
76         if (!dfn[i]) 
77             Tarjan(i);
78     for (int i = 0; i <= ec; ++i)
79         if (col[e[i].u] != col[e[i].v])
80             ++in[col[e[i].v]], ++out[col[e[i].u]];
81     for (int i = 1; i <= color; ++i)
82     
83         if (!in[i]) ++ans;
84         if (!out[i]) ++cnt;
85     
86     printf("%d\n%d", ans, color == 1 ? 0 : max(cnt, ans));
87     return 0;
88 

 

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