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 }