Eddy‘s picture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

Eddy begins to like painting pictures recently ,he is sure of himself to become a painter.Every day Eddy draws pictures in his small room, and he usually puts out his newest pictures to let his friends appreciate. but the result it can be imagined, the friends
are not interested in his picture.Eddy feels very puzzled,in order to change all friends ‘s view to his technical of painting pictures ,so Eddy creates a problem for the his friends of you.

Problem descriptions as follows: Given you some coordinates pionts on a drawing paper, every point links with the ink with the straight line, causes all points finally to link in the same place. How many distants does your duty discover the shortest length
which the ink draws?

Input

The first line contains 0 < n <= 100, the number of point. For each point, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the point.

Input contains multiple test cases. Process to the end of file.

Output

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

Sample Input

3
1.0 1.0
2.0 2.0
2.0 4.0

Sample Output

3.41

Author

eddy

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1162

题意：给你几个点（二维），求将它们连接起来的最小线段的长度。

思路：由于输入的只是点的坐标，所以要先求出它们两两之间的距离。然后就没有然后了。

Prime算法AC代码： 0ms

#include <cstdio>
#include <cmath>
#include <iostream>
using namespace std;
const int INF=0x3f3f3f3f;
struct node
{
    double x,y;
} t[105];
double dis[105];
double a[105][105];
bool vis[105];
int n;
double Prime()
{
    for(int i=0;i<n;i++)
    {
        dis[i]=a[0][i];
        vis[i]=false;
    }
    dis[0]=0;
    vis[0]=true;
    double ans=0;
    for(int i=1;i<n;i++)
    {
        int p=-1;
        double minn=INF;
        for(int j=0;j<n;j++)
        {
            if(!vis[j]&&dis[j]<minn)
                minn=dis[p=j];
        }
        ans+=minn;
        vis[p]=true;
        for(int j=0;j<n;j++)
        {
            if(!vis[j]&&dis[j]>a[p][j])
                dis[j]=a[p][j];
        }
    }
    return ans;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0; i<n; i++)
            scanf("%lf%lf",&t[i].x,&t[i].y);

        for(int i=0; i<n; i++)
        {
            for(int j=0; j<n; j++)
                if(i==j)
                    a[i][j]=0.0;
                else
                    a[i][j]=sqrt((t[i].x-t[j].x)*(t[i].x-t[j].x)+(t[i].y-t[j].y)*(t[i].y-t[j].y));
        }
        printf("%.2lf\n",Prime());
    }
    return 0;
}

Kruskal算法AC代码： 0ms

#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
const int INF=0x3f3f3f3f;

struct Point
{
    double x,y;
}t[100];
struct node
{
    int s,e;
    double w;
}a[5000+5];
int fa[105];
int n,m;
int Find(int x)
{
    if(x==fa[x])
        return x;
    return fa[x]=Find(fa[x]);
}
bool cmp(node a,node b)
{
    return a.w<b.w;
}
double Kruskal()
{
    for(int i=0;i<n;i++)
        fa[i]=i;
    sort(a,a+m,cmp);
    double ans=0;
    for(int i=0;i<m;i++)
    {
        int fx=Find(a[i].s);
        int fy=Find(a[i].e);
        if(fx!=fy)
        {
            ans+=a[i].w;
            fa[fx]=fy;
        }
    }
    return ans;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
            scanf("%lf%lf",&t[i].x,&t[i].y);
        m=0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<i;j++)
            {
                a[m].s=i;
                a[m].e=j;
                a[m++].w=sqrt((t[i].x-t[j].x)*(t[i].x-t[j].x)+(t[i].y-t[j].y)*(t[i].y-t[j].y));
            }
        }
        printf("%.2lf\n",Kruskal());
    }
    return 0;
}

HDU 1162 Eddy's picture【最小生成树，Prime算法+Kruskal算法】