【强连通分量】 HDOJ 4985 Little Pony and Permutation

求一下强连通分量就行。。。

#include <iostream>  #include <queue>  #include <stack>  #include <map>  #include <set>  #include <bitset>  #include <cstdio>  #include <algorithm>  #include <cstring>  #include <climits>  #include <cstdlib>#include <cmath>#include <time.h>#define maxn 100005#define maxm 300005#define eps 1e-10#define mod 1000000007#define INF 999999999#define lowbit(x) (x&(-x))#define mp mark_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid  #define rson o<<1 | 1, mid+1, R#pragma comment(linker, "/STACK:16777216")  typedef long long LL;//typedef int LL;using namespace std;LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}// headint h[maxn], next[maxm], v[maxm];int dfn[maxn], low[maxn], ins[maxn];int p[maxn], cnt, res, top, n, x;stack<int> s;void init(void){cnt = top = 0;memset(h, -1, sizeof h);memset(dfn, 0, sizeof dfn);memset(low, 0, sizeof low);}void addedges(int u, int vv){next[cnt] = h[u], h[u] = cnt, v[cnt] = vv, cnt++;}void tarjan(int u)  {      dfn[u] = low[u] = ++top;      s.push(u), ins[u] = 1;      for(int e = h[u]; ~e; e = next[e]) {          if(!dfn[v[e]]) {              tarjan(v[e]);              low[u] = min(low[u], low[v[e]]);          }          else if(ins[v[e]]) low[u] = min(low[u], dfn[v[e]]);      }      if(dfn[u] == low[u]) {          int tmp = s.top(); s.pop(), ins[tmp] = 0;while(tmp != u) {p[res++] = tmp;tmp = s.top(); s.pop(), ins[tmp] = 0;}p[res++] = tmp;printf("(");for(int i = res-1; i >= 1; i--) printf("%d ", p[i]);printf("%d)", p[0]);    }  }int main(void){while(scanf("%d", &n)!=EOF) {init();for(int i = 1; i <= n; i++) scanf("%d", &x), addedges(i, x);for(int i = 1; i <= n; i++) if(!dfn[i]) res = 0, tarjan(i);printf("\n");}return 0;}