[并查集][最小生成树]PJOJ 2560 雀斑连线/luogu 2921 修复公路

最新推荐文章于 2022-11-05 13:29:47 发布

原创最新推荐文章于 2022-11-05 13:29:47 发布 · 303 阅读

CC 4.0 BY-SA版权

本文介绍了一道关于连接点以形成最小总连线长度的问题，并通过最小生成树算法解决该问题。利用并查集来判断边的两端是否已连接，从而避免形成环路，确保最终形成的树是最小生成树。

题目描述
In an episode of the Dick Van Dyke show, little Richie connects the freckles on his Dad’s back to form a picture of the Liberty Bell. Alas, one of the freckles turns out to be a scar, so his Ripley’s engagement falls through.
Consider Dick’s back to be a plane with freckles at various (x,y) locations. Your job is to tell Richie how to connect the dots so as to minimize the amount of ink used. Richie connects the dots by drawing straight lines between pairs, possibly lifting the pen between lines. When Richie is done there must be a sequence of connected lines from any freckle to any other freckle.
Input

The first line contains 0 < n <= 100, the number of freckles on Dick’s back. For each freckle, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the freckle.
Output

Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the freckles.
Sample Input

3
1.0 1.0
2.0 2.0
2.0 4.0
Sample Output

3.41

分析
en，最小生成树模版题，主要是用并查集判断枚举边的两点是否在同一树内】

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#define rep(i,a,b) for (i=a;i<=b;i++)
using namespace std;
int n,m;
double ans;
double x[101],y[101];
struct G
{
    int x,y;
    double w;
}a[10001];
int f[101];
int i,j;
bool cmp(G a,G b)
{
    return a.w<b.w;
}
int getf(int x)
{
    if (f[x]==x) return x;
    int root=getf(f[x]);
    f[x]=root;
    return root;
}
void add(int x,int y)
{
    int i=getf(x),j=getf(y);
    f[i]=j;
}
int main()
{
    scanf("%d",&n);
    rep(i,1,n)
    f[i]=i;
    rep(i,1,n)
    scanf("%lf%lf",&x[i],&y[i]);
    rep(i,1,n)
    rep(j,i+1,n)
    {
        m++;
        a[m].x=i;a[m].y=j;
        a[m].w=sqrt(pow(x[i]-x[j],2)+pow(y[i]-y[j],2));
    }
    sort(a+1,a+m+1,cmp);
    rep(i,1,m)
    if (getf(a[i].x)!=getf(a[i].y))
    {
        ans+=a[i].w;
        add(a[i].x,a[i].y);
    }
    printf("%.2lf",ans);
}