poj3625 解题报告

来源：互联网发布：淘宝内容营销需求分析编辑：程序博客网时间：2024/06/05 06:04

又是一道最小生成树,拿来测模板;

题意:N个点,坐标给出,然后有M条路已经修好了,给出这M条路的起始点和终点,求需要再修多远的路才能得到最小生成树

题解:把已经修好的路长度算0,然后最小生成树就好了;话说我wa了一次,题目说了全部用64位的运算,我没看到,把坐标改成longlong就过了

好水啊;

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>using namespace std ;const int MAXN = 1005 ;struct  Edge{    int  start , end ;    double length ;} edge[MAXN * MAXN] ;struct  Point{    long long  x , y ;} point[MAXN] ;int  father[MAXN] , Count ;bool vis[MAXN][MAXN] ;double  getlength( Point a , Point b ){    double len ;    len = sqrt( double ( ( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ) ) ) ;    return len ;}bool  cmp( Edge a , Edge b ){    return a.length < b.length ;}int  find( int x ){    if( father[x] == x ) return x ;    father[x] = find( father[x] ) ;    return father[x]              ;}bool Union( int x , int y ){    int  f1 = find( x ) ;    int  f2 = find( y ) ;    if( f1 == f2 ) return false ;    else if( f1 < f2 ) father[f1] = f2 ;    else father[f2] = f1 ;    return true ;}double kruskal( int n ){    int  i , j = 0 ;    double sum = 0 ;    for( i = 0 ; i < n ; i ++ )        father[i] = i ;    sort( edge , edge + Count , cmp ) ;    for( i = 0 ; i < Count && j < n ; i ++ )    {        if( Union( edge[i].start , edge[i].end ) )        {            sum += edge[i].length  ;            j ++ ;        }    }    return sum ;}int  main(){    int  M , N ;    int  i , j , start , end ;    Count = 0 ;    memset( vis , 0 , sizeof ( vis ) ) ;    scanf( "%d%d" , & N , & M ) ;    for( i = 0 ; i < N ; i ++ )        scanf( "%d%d" , & point[i].x , & point[i].y ) ;    for( i = 0 ; i < M ; i ++ )    {        scanf( "%d%d" , & start , & end ) ;        vis[start-1][end-1] = 1 ;        vis[end-1][start-1] = 1 ;    }    Count = 0 ;    for( i = 0 ; i < N - 1 ; i ++ )        for( j = i + 1 ; j < N ; j ++ )        {            edge[Count].start = i ;            edge[Count].end   = j ;            if( vis[i][j] == 0 ) edge[Count].length = getlength( point[i] , point[j] ) ;            else edge[Count].length = 0 ;            Count ++ ;        }    //for( i = 0 ; i < Count ; i ++ )    //    cout << edge[i].start << " " << edge[i].end << " " << edge[i].length << endl ;    printf( "%.2f\n" , kruskal( N ) ) ;    return 0 ;}