题目链接：http://codeforces.com/contest/711/problem/D
题意：n个点n条边的有向图，每个点都指向另外一个点，要翻转其中的几条边，问有多少种翻转方式可以使这张图没有环存在。
可以先求强联通分量，由于题目的特殊性，每个点只有一个出边，所以加入一个强联通分量至少有两个点，则每一个强联通分量中必定有且仅有只有一个环。我们考虑如何翻转这个环内的边，使得生成的数量一共有C(x,1)+C(x,2)+...+C(x,x)=2^x-1种，除去全部翻转的C(x,x)，则还剩下2^x-2种。则所有点数大于2强联通分量内一共有Σ(2^xi-2)种。
再考虑一个出边的强联通分量，无论如何翻转都是没有问题的，则一共可以翻转2^y次。
则总共可以翻转(Σ(2^xi-2))*2^y次。
1 #include <algorithm> 2 #include <iostream> 3 #include <iomanip> 4 #include <cstring> 5 #include <climits> 6 #include <complex> 7 #include <fstream> 8 #include <cassert> 9 #include <cstdio> 10 #include <bitset> 11 #include <vector> 12 #include <deque> 13 #include <queue> 14 #include <stack> 15 #include <ctime> 16 #include <set> 17 #include <map> 18 #include <cmath> 19 using namespace std; 20 #define fr first 21 #define sc second 22 #define cl clear 23 #define BUG puts("here!!!") 24 #define W(a) while(a--) 25 #define pb(a) push_back(a) 26 #define Rint(a) scanf("%d", &a) 27 #define Rll(a) scanf("%I64d", &a) 28 #define Rs(a) scanf("%s", a) 29 #define Cin(a) cin >> a 30 #define FRead() freopen("in", "r", stdin) 31 #define FWrite() freopen("out", "w", stdout) 32 #define Rep(i, len) for(int i = 0; i < (len); i++) 33 #define For(i, a, len) for(int i = (a); i < (len); i++) 34 #define Cls(a) memset((a), 0, sizeof(a)) 35 #define Clr(a, x) memset((a), (x), sizeof(a)) 36 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 37 #define lrt rt << 1 38 #define rrt rt << 1 | 1 39 #define pi 3.14159265359 40 #define RT return 41 #define lowbit(x) x & (-x) 42 #define onecnt(x) __builtin_popcount(x) 43 typedef long long LL; 44 typedef long double LD; 45 typedef unsigned long long ULL; 46 typedef pair<int, int> pii; 47 typedef pair<string, int> psi; 48 typedef pair<LL, LL> pll; 49 typedef map<string, int> msi; 50 typedef vector<int> vi; 51 typedef vector<LL> vl; 52 typedef vector<vl> vvl; 53 typedef vector<bool> vb; 54 55 const int maxn = 200200; 56 const LL mod = 1e9+7; 57 LL ret; 58 59 typedef struct Edge { 60 int u; 61 int v; 62 int next; 63 Edge() { next = -1; } 64 }Edge; 65 66 int head[maxn], ecnt; 67 Edge edge[maxn]; 68 int n, m; 69 70 int bcnt, dindex; 71 int dfn[maxn], low[maxn]; 72 int stk[maxn], top; 73 int belong[maxn]; 74 int in[maxn], out[maxn]; 75 bool instk[maxn]; 76 int cnt[maxn]; 77 78 void init() { 79 memset(edge, 0, sizeof(edge)); 80 memset(head, -1, sizeof(head)); 81 memset(instk, 0, sizeof(instk)); 82 memset(dfn, 0, sizeof(dfn)); 83 memset(low, 0, sizeof(low)); 84 memset(belong, 0, sizeof(belong)); 85 memset(in, 0, sizeof(in)); 86 memset(out, 0, sizeof(out)); 87 ecnt = top = bcnt = dindex = 0; 88 } 89 90 void adde(int uu, int vv) { 91 edge[ecnt].u = uu; 92 edge[ecnt].v = vv; 93 edge[ecnt].next = head[uu]; 94 head[uu] = ecnt++; 95 } 96 97 void tarjan(int u) { 98 int v = u; 99 dfn[u] = low[u] = ++dindex; 100 stk[++top] = u; 101 instk[u] = 1; 102 for(int i = head[u]; ~i; i=edge[i].next) { 103 v = edge[i].v; 104 if(!dfn[v]) { 105 tarjan(v); 106 low[u] = min(low[u], low[v]); 107 } 108 else if(instk[v]) low[u] = min(low[u], dfn[v]); 109 } 110 if(dfn[u] == low[u]) { 111 bcnt++; 112 do { 113 v = stk[top--]; 114 instk[v] = 0; 115 belong[v] = bcnt; 116 cnt[bcnt]++; 117 } while(v != u); 118 } 119 } 120 121 LL quickmul(LL x, LL n) { 122 LL ans = 1; 123 while(n) { 124 if(n & 1) ans = (ans * x) % mod; 125 x = x * x % mod; 126 n >>= 1; 127 } 128 return ans; 129 } 130 131 132 int main() { 133 // FRead(); 134 while(~Rint(n)) { 135 ret = 1; Cls(cnt); 136 init(); 137 int v; 138 For(i, 1, n+1) { 139 Rint(v); 140 adde(i, v); 141 } 142 For(i, 1, n+1) if(!dfn[i]) tarjan(i); 143 LL tmp = 0; 144 For(i, 1, bcnt+1) { 145 if(cnt[i] > 1) { 146 ret = (ret * (quickmul(2, cnt[i])-2+mod))%mod; 147 tmp = (tmp + cnt[i]) % mod; 148 } 149 } 150 ret = (ret * quickmul(2, n-tmp)) % mod; 151 cout << ret << endl; 152 } 153 RT 0; 154 }