Program4_B

本文介绍了一道关于计算几何的问题,即如何通过连线使得n个二维平面上的点彼此连通且连线总长度最短。文章提供了使用最小生成树算法解决此问题的详细步骤和示例代码。

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 我现在做的是第四专题编号为1002的试题,主要内容是:

Problem B

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 38   Accepted Submission(s) : 8
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.<br>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?<br>
 

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

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. <br>
 

Sample Input
3 1.0 1.0 2.0 2.0 2.0 4.0
 

Sample Output
3.41
 
简单题意
给出n个点的坐标(二维x, y),可以使用直线(无向的)将任意两个点连接起来,求将所有点连接起来形成一个整体(使任何两点之间可达),线的最小距离。

解题思路:
最小生成树,用了两点间的距离公式,普利姆算法解决问题

编写代码:

#include<stdio.h>
#include<cstring>
#include<math.h>
double map[105][105];
int v[105];
double lowcost[105];
int n;
void init()
{
    int i,j;
    memset(v,0,sizeof(v));
    for(i=0; i<n; i++)
        for(j=0; j<n; j++)
            map[i][j]=1.7976931348623158e+308;
}
double prim(int u)
{
    int i,j;
    double min,mincost=0;
    v[u]=1;
        for(i=0; i<n; i++)
        lowcost[i]=map[u][i];
    for(i=1; i<n; i++)
    {
        min=1.7976931348623158e+308;
        for(j=0; j<n; j++)
            if(lowcost[j]<min && !v[j])
                min=lowcost[j],u=j;
        v[u]=1;
        mincost+=min;
        for(j=0; j<n; j++)
            if(lowcost[j]>map[u][j] && !v[j])
                lowcost[j]=map[u][j];
    }
    return mincost;
}
int main(){
    int i,j;
    double a[105],b[105];
    while(~scanf("%d",&n) && n) {
        init();
        for(i=0; i<n; i++)
            scanf("%lf%lf",a+i,b+i);
        for(i=0; i<n; i++)
            for(j=i; j<n; j++)
            {
                map[j][i]=map[i][j]=sqrt((a[i]-a[j])*(a[i]-a[j])+(b[i]-b[j])*(b[i]-b[j]));
            }
       printf("%.2f\n",prim(0));
    }
    return 0;
}


### 关于 `program_b` 的介绍 目前,在已有的引用资料中并未提及具体名为 `program_b` 的工具或库。然而,可以推测您可能希望了解类似于 Boost.Program_options 这样的程序选项解析框架的相关内容[^2]。 如果假设 `program_b` 是指某种特定的软件或者工具包,则需要进一步确认其定义范围以及功能描述。通常情况下,类似的命名方式可能是开发者自定义项目的一部分或者是某个开源项目的子模块名称。如果没有明确指向某款知名产品的话,建议从以下几个方面入手: 1. **源码仓库查找**:尝试访问主流代码托管平台(如 GitHub、GitLab),输入关键词 “program_b”,查看是否有匹配的结果。 2. **官方文档查阅**:如果有上下文中提到的具体技术栈背景信息,比如编程语言环境 C++/Python 等,可结合该领域内的常见资源网站进行检索。 3. **社区讨论区提问**:利用 Stack Overflow 或 Reddit 上的技术论坛发布疑问帖寻求解答;同时也可以关注邮件列表公告获取最新动态消息。 以下是基于现有知识体系构建的一个简单示例展示如何实现命令行参数处理逻辑的功能片段供参考: ```cpp #include <boost/program_options.hpp> #include <iostream> namespace po = boost::program_options; int main(int argc, char* argv[]) { try { po::options_description desc("Allowed options"); desc.add_options() ("help", "produce help message") ("optimization,o", po::value<int>()->default_value(1), "set optimization level") ("include-path,I", po::value<std::vector<std::string>>(), "add include path"); po::variables_map vm; store(po::parse_command_line(argc, argv, desc), vm); notify(vm); if (vm.count("help")) { std::cout << desc << "\n"; return 0; } int opt_level = vm["optimization"].as<int>(); const auto& inc_paths = vm["include-path"].as<std::vector<std::string>>(); std::cout << "Optimization level is " << opt_level << '\n'; std::cout << "Include paths are: "; for(const auto &path : inc_paths){ std::cout << path << ' '; } std::cout << '\n'; } catch(std::exception& e) { std::cerr << "error: " << e.what() << "\n"; return 1; } catch(...) { std::cerr << "Exception of unknown type!\n"; return 2; } return 0; } ``` 此段代码展示了使用 Boost.Program_options 库来解析命令行参数的方法,并打印相应的优化级别和包含路径信息。 --- ####
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