洛谷——P2872 [USACO07DEC]道路建设Building Roads

最新推荐文章于 2024-08-03 15:53:55 发布

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最新推荐文章于 2024-08-03 15:53:55 发布

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原文链接：http://www.cnblogs.com/z360/p/7468764.html

本文介绍了一道USACO竞赛题目“道路建设”的解决方案，该问题要求通过计算最小生成树来确定连接所有农场所需的最短额外道路长度。文章提供了完整的C++代码实现，并解释了如何使用距离公式和并查集数据结构。

P2872 [USACO07DEC]道路建设Building Roads

题目描述

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 (Xi, Yi) on the plane (0 ≤ Xi ≤ 1,000,000; 0 ≤ Yi ≤ 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.

给出nn个点的坐标,其中一些点已经连通,现在要把所有点连通,求修路的最小长度.

输入输出格式

输入格式：

Line 1: Two space-separated integers: N and M
Lines 2..N+1: Two space-separated integers: Xi and Yi
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.

输出格式：

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.

输入输出样例

输入样例#1：

输出样例#1：

4.00


最小生成树裸题

#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 510000
using namespace std;
int n,m,x,y,s,fx,fy,xx[N],yy[N],fa[N];
int read()
{
    int x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
struct Edge
{
    int x,y;
    double z;
}edge[N<<1];
int cmp(Edge a,Edge b)
{
    return a.z<b.z;
}
int find(int x)
{
    if(fa[x]==x) return x;
    fa[x]=find(fa[x]);
    return fa[x];
}
int main()
{
    n=read(),m=read();
    for(int i=1;i<=n;i++)
     xx[i]=read(),yy[i]=read();
    for(int i=1;i<=n;i++)
     for(int j=1;j<=n;j++)
      if(i!=j)
      {
           ++s;
           edge[s].x=i;
           edge[s].y=j;
           edge[s].z=sqrt(pow(xx[i]-xx[j],2)+pow(yy[i]-yy[j],2));
      } 
    for(int i=1;i<=n;i++) fa[i]=i;
    double ans=0;
    for(int i=1;i<=m;i++)
    {
        x=read(),y=read();
        fx=find(x),fy=find(y);
        fa[fx]=fy;
    }
    sort(edge+1,edge+1+s,cmp);
    for(int i=1;i<=s;i++)
    {
        x=edge[i].x,y=edge[i].y;
        fx=find(x),fy=find(y);
        if(fa[fx]==fy) continue;
        fa[fx]=fy;
        ans+=edge[i].z;
    }
    printf("%.2lf",ans);
    return 0;
}