Building Roads-优快云博客

本文链接：https://blog.youkuaiyun.com/Mr_JJ_Lian/article/details/6668023

本文介绍了一个使用最小生成树算法解决农场间新建道路问题的方法。通过计算各农场间的距离并运用Prim算法，实现了所有农场之间的最短连接路径。此方案帮助农民John在最少的额外道路建设下，确保任意两个农场都能通过道路相互到达。

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Building Roads

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6364		Accepted: 1831

Description

Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms.

Each of the N (1 ≤ N ≤ 1,000) farms (conveniently numbered 1..N) is represented by a position (X_i, Y_i) on the plane (0 ≤ X_i≤ 1,000,000; 0 ≤ Y_i≤ 1,000,000). Given the preexisting M roads (1 ≤ M ≤ 1,000) as pairs of connected farms, help Farmer John determine the smallest length of additional roads he must build to connect all his farms.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Two space-separated integers: X_iand Y_i
* Lines N+2..N+M+2: Two space-separated integers: i and j, indicating that there is already a road connecting the farm i and farm j.

Output

* Line 1: Smallest length of additional roads required to connect all farms, printed without rounding to two decimal places. Be sure to calculate distances as 64-bit floating point numbers.

Sample Input

Sample Output

4.00

Source

USACO 2007 December Silver

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
double map[1001][1001];
int N,M;
double  prim()
{

 bool vis[1001]={0};
 double lw[1001],min,sum;
 int k;
 for(int i=1;i<=N;i++)
  lw[i]=map[1][i];
 vis[1]=1;
 sum=0;
 for(int i=1;i<N;i++)
 {
  min=10000001;
  for(int j=1;j<=N;j++)
   if(!vis[j] &&  lw[j]<min)
   {
    min=lw[j];
    k=j;
   }
            sum+=min;
   vis[k]=1;
   for(int j=1;j<=N;j++)
    if( !vis[j] && map[k][j]<lw[j])
     lw[j]=map[k][j];
 }
 return sum;
}
double dis(double x1,double y1,double x2,double y2)
{
 return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int main()
{
     double vx[1001],vy[1001],total;
  int n,m;
  while(scanf("%d%d",&N,&M) !=EOF)
  {
   for(int i=1;i<=N;i++)
    scanf("%lf%lf",&vx[i],&vy[i]);
   for(int i=1;i<=N;i++)
    for(int j=i+1;j<=N;j++)
    {
     double temp=dis(vx[i],vy[i],vx[j],vy[j]);
     map[i][j]=map[j][i]=temp;
    }
    for(int i=1;i<=M;i++)
    {
     scanf("%d%d",&n,&m);
     map[n][m]=map[m][n]=0;
    }
    total=prim();
    printf("%.2lf\n",total);
  }

}