POJ 2560 浮点型的带权值

最新推荐文章于 2019-09-20 15:14:29 发布

原创最新推荐文章于 2019-09-20 15:14:29 发布 · 356 阅读

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ACM算法竞赛之路同时被 2 个专栏收录

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最小生成树

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本文介绍了一种解决连接多个点并使总连线长度最短的问题的算法——最小生成树算法。该算法通过寻找一系列的直线连接所有点，同时确保使用的墨水总量最少，适用于诸如地图绘制等实际应用场景。

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Freckles

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8309		Accepted: 3952

Description

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

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct edge
{
    int x,y;
    float w;
}e[1000000];
int f[1000000];
float x[1000];
float y[1000];
int cmp(edge a,edge b)
{
    return a.w<b.w;
}
int find(int x)
{
    return f[x] == x ? x : (f[x] = find(f[x])); //并查集一步写完
}
int merge(int a,int b)
{
    int u,v;
    u=find(a);
    v=find(b);
    if(u!=v){
    f[v]=u;
    return 1;
}
return 0;
}
int main()
{
    int n;
    int cnt=0;
    scanf("%d",&n);
    for(int i=0;i<1000000;i++)
    {
        f[i]=i;
    }
    for(int i=0;i<n;i++)
    {
        scanf("%f%f",&x[i],&y[i]);
        for(int j=0;j<i;j++)
        {
            e[cnt].x=i;
            e[cnt].y=j;
            e[cnt].w=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
            cnt++;
        }
    }
    float ans=0;
    sort(e,e+cnt,cmp);
    for(int i=0;i<cnt;i++)
    {


           if(merge(e[i].x,e[i].y))
             ans+=e[i].w;
        }

    printf("%.2f",ans);

return 0;
}