zoukankan      html  css  js  c++  java
  • ZOJ 1872 Freckles



    Freckles

    Time Limit: 2 Seconds      Memory Limit: 65536 KB

    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.

    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 freckles.


    Sample Input

    3
    1.0 1.0
    2.0 2.0
    2.0 4.0


    Sample Output

    3.41


    Source: University of Waterloo Local Contest 2000.09.23


    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>

    using namespace std;
    #define CLS(x,c) memset(x,c,sizeof(x))
    const int maxn=1111;
    int n;
    struct node
    {
        double x,y;
    }p[111];
    double gp[maxn][maxn];
    double distan(node a,node b)
    {
        return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
    }

    double prim(int n)
    {
        double lowcast[maxn];
        bool s[maxn];

        for(int i=1;i<=n;i++)
        {
            lowcast=distan(p[1],p);
            s=0;
        }

        s[1]=1;double sum=0;
        for(int i=2;i<=n;i++)
        {
            double minn=0x3f3f3f3f;
            int u=-1;
            for(int j=1;j<=n;j++)
            {
                if(!s[j]&&lowcast[j]<minn)
                {
                    minn=lowcast[j];
                    u=j;
                }
            }
            if(u==-1) return 0x3f3f3f3f;
            sum+=minn; s=1;

            for(int j=1;j<=n;j++)
            {
                if(!s[j]&&distan(p,p[j])<lowcast[j])
                    lowcast[j]=distan(p,p[j]);
            }
        }

        return sum;
    }

    int main()
    {
    while(scanf("%d",&n)!=EOF)
    {
        CLS(gp,0);
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf",&p.x,&p.y);
        }
        printf("%.2lf ",prim(n));
    }
        return 0;
    }

  • 相关阅读:
    github常见操作和常见错误!错误提示:fatal: remote origin already exists.
    [vim]vim 在win下乱码解决
    [git]图解git常用命令
    [vim配置]windows下在vim中使用gcc/g++编译调试c/cpp文件
    windows live writer 尝试登陆时发生意外错误,导致无法发表博客解决方案
    win7系统cmd命令切换到指定文件夹目录
    qt中文乱码问题
    免费Flash图表工具FusionChart
    将matlab中数据输出保存为txt或dat格式
    对接支付宝的城市服务(水费)
  • 原文地址:https://www.cnblogs.com/CKboss/p/3350923.html
Copyright © 2011-2022 走看看