畅通工程再续 hdu1875-优快云博客

本文链接：https://blog.youkuaiyun.com/huixisheng/article/details/5408717

本文详细解析了使用最小生成树算法解决特定问题的过程，并通过具体的编程实现进行了验证。重点介绍了如何利用距离比较进行节点间连接的选择，以确保最终形成的网络满足特定条件。此外，还对比了不同实现方式对算法效率的影响。

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http://acm.hdu.edu.cn/showproblem.php?pid=1875

#include<iostream>//2235167 2010-03-23 16:21:03 Accepted 1875 312MS 352K 1567 B C++ 悔惜晟
#include<algorithm> //效率真的不高，算法还是有待给进,耗时主要在哪里呢？？
#include<cstdio>
#include<cmath>
int aa[4951];
using namespace std;

struct stu1
{
  int x1;
  int y1;
}df1[103];

struct stu2
{
  int x2;
  int y2;
  double distance ;
}df2[4951];

int cmp(stu2 a, stu2 b)
{
  return a.distance < b.distance;
}

int find(int x)
{
  int r = x;
  while(r !=aa[r])
   r = aa[r];
  return r;
}

void mergy(int a, int b)
{
  if(a > b)
   aa[b] = a;
  else
   aa[a] = b;
}

int main()
{
int t, n, i, j, count, ww;
double sum, dis;
scanf("%d", &t);
while(t--)
{
  scanf("%d", &n);
  for(i = 0; i < n; i++)
   scanf("%d%d", &df1[i].x1, &df1[i].y1);
   count = 0;
  for(i = 0 ; i < n-1; i++)
   for(j = i+1; j < n; j++)
   {

     dis = sqrt(1.0*((df1[i].x1 - df1[j].x1) * (df1[i].x1 - df1[j].x1) + (df1[i].y1 - df1[j].y1) * (df1[i].y1 - df1[j].y1)));
         if(dis < 10 || dis > 1000)
           continue;
      else
         {
         df2[count].x2 = i + 1;
      df2[count].y2 = j + 1;
      df2[count].distance = dis;
      count++;
     }

   }

   sort(df2,df2+count,cmp);
   for(i = 1 ; i <= n*(n-1)/2; i++)
    aa[i] = i;
   ww = 0;
   sum = 0;//忘记了初始话
   for(i = 0 ;i < count ; i++)
   {
       int a = find(df2[i].x2);
       int b = find(df2[i].y2);
       if(a != b)
       {
       sum += df2[i].distance;
       mergy(a, b);
       ww++;
    }
   }
   if(ww == n-1)
       printf("%0.1lf/n", 100*sum);
   else
                printf("oh!/n");
}

}

/*
#include<iostream>//2235212 2010-03-23 16:25:50 Time Limit Exceeded 1875 1000MS 408K 1648 B C++ 悔惜晟
#include<algorithm>
#include<cstdio>
#include<cmath>
int aa[9910];
using namespace std;

struct stu1
{
  int x1;
  int y1;
}df1[103];

struct stu2
{
  int x2;
  int y2;
  double distance ;
}df2[9910];

int cmp(stu2 a, stu2 b)
{
  return a.distance < b.distance;
}

int find(int x)
{
  int r = x;
  while(r !=aa[r])
   r = aa[r];
  return r;
}

void mergy(int a, int b)
{
  if(a > b)
   aa[b] = a;
  else
   aa[a] = b;
}

int main()
{
int t, n, i, j, num, count, flag;
double sum, dis;
scanf("%d", &t);
while(t--)
{
  scanf("%d", &n);
  for(i = 0; i < n; i++)
  scanf("%d%d", &df1[i].x1, &df1[i].y1);
  count = 0;
  flag = 0;
  for(i = 0 ; i < n - 1; i++)
  {
   int tt = 0;
   for(j = 0; j < n; j++)
   {
    if(i != j)//这里处理的很不好，多做很多没有用的判断
    {
     dis = sqrt(1.0*((df1[i].x1 - df1[j].x1) * (df1[i].x1 - df1[j].x1) + (df1[i].y1 - df1[j].y1) * (df1[i].y1 - df1[j].y1)));
         if(dis < 10 || dis > 1000)
           t++;
      else
         {
         df2[count].x2 = i + 1;
      df2[count].y2 = j + 1;
      df2[count].distance = dis;
      count++;
     }
    }

   }
   if(tt == n-1)
   {
    flag = 1;
    break;
   }
  }
  if( flag == 1)
   printf("oh!/n");
  else
  {
   sort(df2,df2+count,cmp);
   for(i = 1 ; i <= n*(n-1)/2; i++)
    aa[i] = i;
   sum = 0;
   for(i = 0 ;i < count ; i++)
   {
       int a = find(df2[i].x2);
       int b = find(df2[i].y2);
       if(a != b)
       {
       sum += df2[i].distance;
       mergy(a, b);
    }
   }
   printf("%0.1lf/n", 100*sum);
  }

}
}