BZOJ-1050: [HAOI2006]旅行comf

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=1050

枚举最大最小边，然后并查集维护连通性即可。

代码：

#include 

#include 

#include 

 

using namespace std ;

 

#define inf 0x7fffffff

#define MAXN 510

#define MAXM 5010

 

struct Edge {

    int s , t , d ;

    bool operator < ( const Edge &a ) const {

        return d < a.d ;

    }

} e[ MAXM ] ;

 

int n , m , S , T ;

 

struct Uset {

     

    int father[ MAXN ] , size[ MAXN ] ;

     

    void Init(  ) {

        memset( father , 0 , sizeof( father ) ) ;

        for ( int i = 0 ; i < MAXN ; ++ i ) size[ i ] = 1 ;

    }

     

    int Find( int x ) {

        int i , j = x ;

        for ( i = x ; father[ i ] ; i = father[ i ] ) ;

        while ( father[ j ] ) {

            int k = father[ j ] ; father[ j ] = i ; j = k ;

        }

        return i ;

    }

     

    bool check( int x , int y ) {

        return Find( x ) == Find( y ) ;

    }

     

    void Union( int x , int y ) {

        if ( Find( x ) != Find( y ) ) {

            if ( size[ Find( x ) ] < size[ Find( y ) ] ) swap( x , y ) ;

            size[ Find( x ) ] += size[ Find( y ) ] ;

            father[ Find( y ) ] = Find( x ) ;

        }

    }

     

} us ;

 

int gcd( int x , int y ) {

    if ( x < y ) swap( x , y ) ;

    while ( y ) {

        int k = x % y ;

        x = y ; y = k ;

    }

    return x ;

}

 

int main(  ) {

    scanf( "%d%d" , &n , &m ) ;

    us.Init(  ) ;

    for ( int i = 0 ; i ++ < m ; ) {

        scanf( "%d%d%d" , &e[ i ].s , &e[ i ].t , &e[ i ].d ) ;

        us.Union( e[ i ].s , e[ i ].t ) ;

    }

    scanf( "%d%d" , &S , &T ) ;

    if ( ! us.check( S , T ) ) {

        printf( "IMPOSSIBLE\n" ) ; return 0 ;

    }

    sort( e + 1 , e + m + 1 ) ;

    int rec = inf , ret = 1 ;

    for ( int i = 0 ; i ++ < m ; ) {

        us.Init(  ) ;

        for ( int j = i ; j <= m ; ++ j ) {

            us.Union( e[ j ].s , e[ j ].t ) ;

            if ( us.check( S , T ) ) {

                if ( double( rec ) / double( ret ) > double( e[ j ].d ) / double( e[ i ].d ) ) {

                    rec = e[ j ].d , ret = e[ i ].d ;

                }

                break ;

            }

        }

    }

    if ( ! ( rec % ret ) ) printf( "%d\n" , rec / ret )

    ; else printf( "%d/%d\n" , rec / gcd( rec, ret ) , ret / gcd( rec , ret ) ) ;

    return 0 ;

}