• hdu 1875 给出每个结点的坐标 权值为两点间的距离 (MST)


    Sample Input
    2
    2
    10 10 //坐标
    20 20
    3
    1 1
    2 2
    1000 1000

    Sample Output
    1414.2   //最小权值和*100  保留1位小数
    oh!       //不连通

    prim

     1 # include <iostream>
     2 # include <cstdio>
     3 # include <cstring>
     4 # include <algorithm>
     5 # include <cmath>
     6 # define LL long long
     7 using namespace std ;
     8 
     9 const int INF=0x3f3f3f3f;
    10 const int MAXN=110;
    11 bool vis[MAXN];
    12 double lowc[MAXN];
    13 int n ;
    14 double cost[MAXN][MAXN] ;
    15 
    16 struct poin
    17 {
    18     int x ;
    19     int y ;
    20 }p[110];
    21 
    22 double Prim()//点是0~n-1
    23 {
    24     double ans=0;
    25     memset(vis,false,sizeof(vis));
    26     vis[0]=true;
    27     for(int i=1;i<n;i++)lowc[i]=cost[0][i];
    28     for(int i=1;i<n;i++)
    29     {
    30         double minc=INF;
    31         int p=-1;
    32         for(int j=0;j<n;j++)
    33             if(!vis[j]&&minc>lowc[j])
    34             {
    35                 minc=lowc[j];
    36                 p=j;
    37             }
    38             if(minc==INF)return -1;//原图不连通
    39             ans+=minc;
    40             vis[p]=true;
    41             for(int j=0;j<n;j++)
    42                 if(!vis[j]&&lowc[j]>cost[p][j])
    43                     lowc[j]=cost[p][j];
    44     }
    45     return ans;
    46 }
    47 
    48 int main()
    49 {
    50 
    51    // freopen("in.txt","r",stdin) ;
    52     int T ;
    53     scanf("%d" , &T) ;
    54     while(T--)
    55     {
    56         scanf("%d" , &n) ;
    57         int i , j ;
    58         for (i = 0 ; i < n ; i++)
    59             for (j = 0 ; j < n ; j++)
    60                cost[i][j] = INF ;
    61         for (i = 0 ; i < n ; i++)
    62             scanf("%d %d" , &p[i].x , &p[i].y) ;
    63         for (i = 0 ; i < n ; i++)
    64             for (j = i+1 ; j < n ; j++)
    65             {
    66                 double t = sqrt((double)(p[i].x - p[j].x) * (p[i].x - p[j].x) + (p[i].y - p[j].y) * (p[i].y - p[j].y)) ;
    67                 if (t <10.0 || t > 1000.0)
    68                     continue ;
    69                 cost[i][j] = t ;
    70                 cost[j][i] = t ;
    71             }
    72         double k = Prim() ;
    73         if (k == -1)
    74             printf("oh!
    ") ;
    75         else
    76             printf("%.1lf
    " , k*100) ;
    77 
    78     }
    79     return 0 ;
    80 }
    View Code
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  • 原文地址:https://www.cnblogs.com/mengchunchen/p/4575163.html
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