Tarjan UVALive 6511 Term Project

题目传送门

 1 /*
 2     题意：第i个人选择第a[i]个人，问组成强联通分量(自己连自己也算)外还有多少零散的人
 3     有向图强联通分量-Tarjan算法：在模板上加一个num数组，记录每个连通分量的点数，若超过1，则将连通点数相加
 4         用总点数-ans则是零散的点
 5     详细解释：http://www.bubuko.com/infodetail-927304.html
 6 */
 7 #include <cstdio>
 8 #include <cstring>
 9 #include <algorithm>
10 #include <vector>
11 #include <stack>
12 using namespace std;
13
14 const int MAXN = 1e5 + 10;
15 const int INF = 0x3f3f3f3f;
16 vector<int> G[MAXN];
17 stack<int> S;
18 int pre[MAXN], low[MAXN];
19 int instack[MAXN], num[MAXN];
20 int dfs_clock, scc_cnt;
21 int n, ans;
22
23 void DFS(int u)
24 {
25     instack[u] = 1;
26     pre[u] = low[u] = ++dfs_clock;
27     S.push (u);
28     for (int i=0; i<G[u].size (); ++i)
29     {
30         int v = G[u][i];
31         if (!pre[v])
32         {
33             DFS (v);    low[u] = min (low[u], low[v]);
34         }
35         else if (instack[v] == 1)
36         {
37             low[u] = min (low[u], pre[v]);
38         }
39     }
40
41     if (low[u] == pre[u])
42     {
43         scc_cnt++;
44         for (; ; )
45         {
46             int x = S.top ();    S.pop ();
47             instack[x] = 0;
48             num[scc_cnt]++;
49             if (x == u)    break;
50         }
51     }
52 }
53
54 void Tarjan(void)
55 {
56     dfs_clock = scc_cnt = 0;
57     memset (pre, 0, sizeof (pre));
58     memset (low, 0, sizeof (low));
59     memset (num, 0, sizeof (num));
60     memset (instack, 0, sizeof (instack));
61     while (!S.empty ())    S.pop ();
62
63     for (int i=1; i<=n; ++i)
64     {
65         if (!pre[i])    DFS (i);
66     }
67 }
68
69 int main(void)        //UVALive 6511 Term Project
70 {
71     freopen ("L.in", "r", stdin);
72
73     int t;    scanf ("%d", &t);
74     while (t--)
75     {
76         ans = 0;    scanf ("%d", &n);
77         for (int i=1; i<=n; ++i)    G[i].clear ();
78         for (int i=1; i<=n; ++i)
79         {
80             int x;    scanf ("%d", &x);
81             G[i].push_back (x);
82             if (i == x)    ans++;
83         }
84
85         Tarjan ();
86         for (int i=1; i<=scc_cnt; ++i)
87         {
88             if (num[i] > 1)    ans += num[i];
89         }
90         printf ("%d\n", n - ans);
91     }
92
93     return 0;
94 }