CodeForces 711D Directed Roads (DFS判环+计数)

Posted 2020-08-05 dwtfukgv

题意：给定一个有向图，然后你可能改变某一些边的方向，然后就形成一种新图，让你求最多有多少种无环图。

析：假设这个图中没有环，那么有多少种呢？也就是说每一边都有两种放法，一共有2^x种，x是边数，那么如果有环呢？假设x是这个连通块的边数，

y是这个环的边数，那么就一共有2^x * (2 ^ y - 2) 种，减去这两种就是一边也不变，和所有的边都就变，这样就形成环了。

那么怎么计算呢？这个题的边很特殊，所以我们可以利用这个特征，我们在每个连通块时，用vis记录一下开始的父结点，用cnt记录每一个到每个点的深度，

然后如果产生环了，那么我们就可以很轻松的算出这个环的结点数，用当前的cnt减去就好，然后用sum记录一下环结点的总数，

最后用n减去环中的结点数，就剩下不是环的结点数了。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <list>
#include <sstream>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e5 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int vis[maxn];
int a[maxn], cnt[maxn];
LL ans;
int sum;

LL quick_pow(LL a, LL b){
    LL ans = 1;
    while(b){
        if(b & 1)  ans = (ans * a) % mod;
        b >>= 1;
        a = (a * a) % mod;
    }
    return ans;
}

void dfs(int d, int u, int fa){
    vis[u] = fa;
    cnt[u] = d;
    if(!vis[a[u]])  dfs(d+1, a[u], fa);
    else if(vis[a[u]] == fa){
        sum += cnt[u]-cnt[a[u]]+1;
        ans = (ans * (quick_pow(2LL, cnt[u]-cnt[a[u]]+1) - 2 + mod)) % mod;
    }
}

int main(){
    while(scanf("%d", &n) == 1){
        memset(vis, 0, sizeof vis);
        memset(cnt, 0, sizeof cnt);
        for(int i = 1; i <= n; ++i)  scanf("%d", &a[i]);

        ans = 1, sum = 0;
        for(int i = 1; i <= n; ++i){
            if(vis[i])  continue;
            dfs(0, i, i);
        }

        ans = (ans * quick_pow(2LL, n-sum)) % mod;
        cout << ans << endl;
    }
    return 0;
}