【算法】带权完全图中的最大权完全子图

Eric_Xu_0925 2025-06-27 16:09:06

求助:

假设带权完全图中有m个顶点,每两个顶点间边的权重已知,

如何在这个完全图中找到一个有n个顶点的完全子图(n≤m),使该子图中边的权重和最大?

换个问法:如何选取n个顶点,使这n个顶点构成的完全图的边的权重和最大?

暴力算法的复杂度为Cmn(阶乘),是否有更快速的算法?

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"一、实验目的和要求 " "(1)掌握图的相关概念,包括图,有向图,无向图,完全子图,连通图,度,入 " "度,出度,简单回路和环等定义。 " "(2)重点掌握图的各种存储结构,包括邻接矩阵和邻接表等。 " "(3)重点掌握图的基本运算,包括创建图,输出图,深度优先遍历,广度优先遍历等 " "。 " "(4)掌握图的其他运算 ,包括最小生成树,最短路径,拓扑排序和关键路径等算法。" "(5)灵活运用图这种数据结构解决一些综合应用问题。 " "二、实验内容和方法 " "(1)实验内容: " "1、编写一个程序algo8-1.cpp,实现不带权图和带权图的邻接矩阵与邻接表的相互转换" "算法、输出邻接矩阵与邻接表的算法,并在此基础上设计一个程序exp8-1.cpp实现如下" "功能: " " 建立如图1所示的有向图G的邻接矩阵,并输出; " " 由有向图G的邻接矩阵产生邻接表,并输出; " " 再由 的邻接表产生对应的邻接矩阵,并输出。 " " " " " " " " " " " " " " " " " "图1 " " " " " "2、编写一个程序algo8-2.cpp,实现图的遍历运算,并在此基础上设计一个程序exp8-2" ".cpp完成如下功能: " " 输出图1所示的有向图G从顶点0开始的深度优先遍历序列(递归算法); " " 输出图1所示的有向图G从顶点0开始的深度优先遍历序列(非递归算法); " " 输出图1所示的有向图G从顶点0开始的广度优先遍历序列。 " " " "3、设计一个程序exp8-3.cpp,采用邻接表存储图,并输出图8.1(a)从指定顶点1出 " "发的所有深度优先遍历序列。 " " " "(2)实验方法: " "1、综合运用课本所学的知识,用不同的算法实现在不同的程序功能。 " "2、结合指导老师的指导,解决程序的问题,正确解决实际存在的异常情况,逐步 " "改善功能。 " "3、根据实验内容,编译程序。 " " " "三、实验环境: " "Windows 7,Visual C++6.0 " "三、实验过程描述 " "文件graph.h定义了图的邻接矩阵表示类型和邻接表表示类型,该头文件在以下三个 " "实验都会使用到。其代码如下: " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " "实验 " "源程序。 " "一、输入如下所示程序; " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " "二、编译并链接程序; " " " "三、运行程序,结果如下图: " " " " " "实验 " "源程序 " "一、输入如下所示程序; " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " "

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