ZROI#1006

Posted 2022-11-26 equinox-flower

ZROI#1006

可能一眼看起来是个很不可做的题,但你仔细思考一下,你发现.....给的是个\(n\)个点\(n\)条边的东西...
那么它可能是个啥呢?
是个基环树?是个森林+环?是个基环树+森林?
都是有可能的!
然后我们发现,答案就是连通块个数+环数-1.
为什么呢?
假设有\(x\)个连通块,\(y\)个环,那么对于连通块我们一定有一个作为答案连通块,其余的都要接在这下面,且只需修改根即可,所以连通块对答案的贡献是\(x-1\),而对于环,必须拆开,且只需要断开一处成为链接在答案连通块之下即可,所以环的贡献是\(y\).

可得答案即为\(x+y-1\).

\(Code:\)

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i)
#define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i)
#define pii pair < int , int >
#define one first
#define two second
#define rint read<int>
#define pb push_back

using std::queue ;
using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () 
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) 
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        
       while ( ch >= '0' && ch <= '9' ) 
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       
   return f * x ;


const int N = 2e5 + 100 ;
struct edge  int to , next ;  e[N] ;
int head[N] , tot ;
int n , f[N] , ans , cnt , in[N] , siz = 0 ;
queue < int > q ; bool vis[N] , founded ;

inline void build (int u , int v) 
    e[++tot].next = head[u] ;
    head[u] = tot ; e[tot].to = v ;


inline void topsort () 
    rep ( i , 1 , n ) if ( ! in[i] ) q.push ( i ) ;
    while ( ! q.empty () ) 
        int j = q.front () ; q.pop () ; vis[j] = true ;
        for (int i = head[j] ; i ; i = e[i].next)
         int k = e[i].to ; -- in[k] ; if ( ! in[k] ) q.push ( k ) ; 
    
    return ;


inline void dfs (int cur) 
    vis[cur] = true ; ++ siz ;
    for (int i = head[cur] ; i ; i = e[i].next)
        if ( ! vis[e[i].to] ) dfs ( e[i].to ) ;
    return ;


int main (int argc , char * argv[]) 
    n = rint () ;
    rep ( i , 1 , n ) 
        f[i] = rint () ;
        build ( i , f[i] ) ;
        ++ in[f[i]] ;
    
    topsort () ;
    rep ( i , 1 , n )
        if ( ! vis[i] ) 
            siz = 0 ; dfs ( i ) ;
            if ( siz == 1 && ! founded ) founded ^= 1 ;
            else ++ ans ;
        
    printf ("%d\n" , ans ) ;
    return 0 ;