Codeforces Round #369 (Div. 2) D 欧拉回路

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题目传送门： http://codeforces.com/contest/711/problem/D

题意：一张图，给n个数a1,a2,.....,an，代表从i到ai，这样建一张图，问有多少种不同的集合，改变集合里的边使图不含环

思路：求出图中环的数量ｎ，每个环中元素的个数ｍ，不在环中元素的个数ｋ，答案 ans = (2^m1-2) * (2^m2-2)*......*(2^mn-2) * (2^k)

代码如下：

#include <iostream>#include <algorithm>#include <cstring>#include <stdio.h>#include <string>#include <cmath>#include <queue>#include <set>#include <map>#include <stack>using namespace std;#define   lson          l,m,rt<<1#define   rson          m+1,r,rt<<1|1#define   ll            long long#define   ull           unsigned long long#define   mem(n,v)      memset(n,v,sizeof(n))#define   MAX           200005#define   MAXN          3000#define   PI            3.1415926#define   E             2.718281828459#define   opnin         freopen("text.in.txt","r",stdin)#define   opnout        freopen("text.out.txt","w",stdout)#define   clsin         fclose(stdin)#define   clsout        fclose(stdout)#define   haha1         cout << "haha1"<< endl#define   haha2         cout << "haha2"<< endl#define   haha3         cout << "haha3"<< endlconst int    INF    =   0x3f3f3f3f;const ll     INFF   =   0x3f3f3f3f3f3f3f3f;const double pi     =   3.141592653589793;const double inf    =   1e18;const double eps    =   1e-8;const ll     mod    =   1e9+7;const ull    mx     =   133333331;/**************************************************************************/int vis[MAX];int to[MAX];ll mul[MAX];int dfs(int u,int id){    vis[u] = id;    if(!vis[to[u]]) return dfs(to[u],id);    else return to[u];}int dfs1(int u,int id){    if(to[u] == id) return 1;    else return dfs1(to[u],id) + 1;}void getmul(int n){    mul[0] = 1;    for(int i=1;i<=n;i++)        mul[i] = (mul[i-1] * 2) % mod;}int main(){    mem(vis,0);    mem(to,0);    mem(mul,0);    int n;    cin >> n;    getmul(n);    for(int i=1;i<=n;i++){        scanf("%d",&to[i]);    }    ll ans = 1      ll sum = 0;    for(int i=1;i<=n;i++){        int cnt = 0;        if(!vis[i]){            int x = dfs(i,i);            if(vis[x] == i) cnt = dfs1(x,x);        }        if(cnt > 1){            ans = (ans * ((mul[cnt] - 2 + mod) % mod)) % mod;            sum += cnt;        }    }    ans = (ans*mul[n-sum])%mod;    cout << ans << endl;    return 0;}

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