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

转载最新推荐文章于 2024-08-03 15:53:55 发布 · 101 阅读

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原文链接：http://www.cnblogs.com/ruojisun/p/7416690.html

本文介绍了一个使用Kruskal算法解决农场道路连接问题的例子。通过计算各个农场之间的最短距离并采用最小生成树的方法，实现了所有农场间的最优连接路径。具体包括输入输出格式及算法实现代码。

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

裸kruskal

屠龙宝刀点击就送

#include <algorithm>
#include <cstdio>
#include <cmath>
#define N 1000005
typedef long long LL;
using namespace std;
int    cnt,fa[1005],n,m,q;
LL x[1005],y[1005];
struct Edge
{
    int x,y;
    double dist;
    bool operator<(Edge a)const
    {
        return dist<a.dist;
    }
}edge[N];
int find_(int x) {return x==fa[x]?x:fa[x]=find_(fa[x]);}
double calc(LL x1,LL y1,LL x2,LL y2) {return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i)
    {
        fa[i]=i;
        scanf("%lld%lld",&x[i],&y[i]);
    }
    for(int u,v,i=1;i<=m;++i)
    {
        scanf("%d%d",&u,&v);
        fa[find_(v)]=find_(u);
    }
    for(int i=1;i<=n;++i)
     for(int j=i+1;j<=n;++j)
      edge[++cnt]=(Edge){i,j,calc(x[i],y[i],x[j],y[j])};
    sort(edge+1,edge+1+cnt);
    double sum=0;
    for(int num=0,i=1;i<=cnt;++i)
    {
        int fx=find_(edge[i].x),fy=find_(edge[i].y);
        if(fx!=fy)
        {
            fa[fy]=fx;
            sum+=edge[i].dist;
            if(++num==n-1) break;
        }
    }
    printf("%.2lf",sum);
    return 0;
}

转载于:https://www.cnblogs.com/ruojisun/p/7416690.html