hdu 1162 裸最下生成树 prim

来源：互联网发布：高中历史辅助教程知乎编辑：程序博客网时间：2024/05/16 10:39

题意：不解释，很简单。

思路：刚刚开始准备用克鲁斯卡尔算法，但一想这是稠密图，故用prim算法解之，看下面代码吧，好久没写代码了，准备各种考试，但是各种不顺，还是写代码爽，哈哈。。

代码：

#include <stdio.h>
#include <string.h>
#include <math.h>

#define N 105
#define INF 1 << 30

struct point
{
   double x, y;
}a[N];

int main()
{
   int i, j, k, n, v;
   double min, sum;
   double edge[N][N], dist[N], near[N];
   while(scanf("%d", &n) != EOF){
       memset(near, 0, sizeof(near));
       for(i = 1; i <= n; ++i){
           scanf("%lf %lf", &a[i].x, &a[i].y);
       }
       for(i = 1; i < n; ++i){
           for(j = i + 1; j <= n; ++j){
               edge[i][j] = edge[j][i] = (a[i].x - a[j].x) * (a[i].x - a[j].x)
                           + (a[i].y - a[j].y) * (a[i].y - a[j].y);
           }
       }
       for(i = 1; i <= n; ++i){
           dist[i] = edge[1][i];
       }
       near[1] = -1;
       sum = 0;
       for(i = 1; i < n; ++i){
           min = INF; v = -1;
           for(j = 1; j <= n; ++j){
               if(near[j] != -1 && dist[j] < min)
                   min = dist[j], v = j;
           }
           if(v != -1){
               sum += sqrt(min);
               near[v] = -1;
               for(j = 1; j <= n; ++j){
                   if(near[j] != -1 && edge[v][j] < dist[j]){
                       dist[j] = edge[v][j];
                   }
               }
           }
       }
       printf("%.2lf\n", sum);
   }
   return 0;
}