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  • [BOI2003]团伙

    [BOI2003]团伙

    也是并查集的扩展域写法.

    这里的域相对于食物链要少一个,因为只有两种关系:敌人,朋友.

    唯一需要注意的是 (:) 朋友的敌人没说是敌人 (...)

    然后就没了...

    (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
    #define db double
    #define ull unsigned long long
    #define lowbit(x) ( x & ( - x ) )
    
    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 = 1e3 + 100 ;
    
    int n , m , f[N*2] ;
    
    inline int getf (int x) { return f[x] == x ? x : f[x] = getf ( f[x] ) ; }
    
    inline void merge (int x , int y) {
        x = getf ( x ) ; y = getf ( y ) ;
        if ( x != y ) f[y] = x ; return ;
    }
    
    signed main (int argc , char * argv[]) {
        n = rint () ; m = rint () ;
        rep ( i , 1 , n * 2 ) f[i] = i ;
        char opt[2] ; int u , v ;
        while ( m -- ) {
            scanf ("%s" , opt ) ;
            u = rint () ; v = rint () ;
            if ( opt[0] == 'E' ) {
                merge ( u , v + n ) ;
                merge ( v , u + n ) ;
            } else merge ( u , v ) ;
        }
        rep ( i , 1 , n ) if ( f[i] == i ) ++ f[0] ;
        printf ("%d
    " , f[0] ) ;
        #ifndef ONLINE_JUDGE
        system ("pause") ;
        #endif
        return 0 ;
    }
    
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  • 原文地址:https://www.cnblogs.com/Equinox-Flower/p/11773806.html
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