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算法系列15天速成――第十五天 图【下】(大结局)

2024-09-08 23:18:41
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今天是大结局,说下“图”的最后一点东西,“最小生成树“和”最短路径“。

一: 最小生成树

1. 概念

    首先看如下图,不知道大家能总结点什么。

    对于一个连通图G,如果其全部顶点和一部分边构成一个子图G1,当G1满足:

       ① 刚好将图中所有顶点连通。②顶点不存在回路。则称G1就是G的“生成树”。

           其实一句话总结就是:生成树是将原图的全部顶点以最小的边连通的子图,这不,如下的连通图可以得到下面的两个生成树。

       ② 对于一个带权的连通图,当生成的树不同,各边上的权值总和也不同,如果某个生成树的权值最小,则它就是“最小生成树”。

     

2. 场景

      实际应用中“最小生成树”还是蛮有实际价值的,教科书上都有这么一句话,若用图来表示一个交通系统,每一个顶点代表一个城市,

  边代表两个城市之间的距离,当有n个城市时,可能会有n(n-1)/2条边,那么怎么选择(n-1)条边来使城市之间的总距离最小,其实它

  的抽象模型就是求“最小生成树”的问题。

 

3. prim算法

    当然如何求“最小生成树”问题,前人都已经给我们总结好了,我们只要照葫芦画瓢就是了,

    第一步:我们建立集合“V,U",将图中的所有顶点全部灌到V集合中,U集合初始为空。

    第二步: 我们将V1放入U集合中并将V1顶点标记为已访问。此时:U(V1)。

    第三步: 我们寻找V1的邻接点(V2,V3,V5),权值中发现(V1,V2)之间的权值最小,此时我们将V2放入U集合中并标记V2为已访问,

                此时为U(V1,V2)。

    第四步: 我们找U集合中的V1和V2的邻接边,一阵痉挛后,发现(V1,V5)的权值最小,此时将V5加入到U集合并标记为已访问,此时

                 U的集合元素为(V1,V2,V5)。

    第五步:此时我们以(V1,V2,V5)为基准向四周寻找最小权值的邻接边,发现(V5,V4)的权值最小,此时将V4加入到U集合并标记

                 为已访问,此时U的集合元素为(V1,V2,V5,V4)。

    第六步: 跟第五步形式一样,找到了(V1,V3)的权值最小,将V3加入到U集合中并标记为已访问,最终U的元素为(V1,V2,V5,V4,V3),

最终发现顶点全部被访问,最小生成树就此诞生。

复制代码 代码如下:

#region prim算法获取最小生成树
        /// <summary>
/// prim算法获取最小生成树
/// </summary>
/// <param name="graph"></param>
        public void Prim(MatrixGraph graph, out int sum)
        {
            //已访问过的标志
            int used = 0;

            //非邻接顶点标志
            int noadj = -1;

            //定义一个输出总权值的变量
            sum = 0;

            //临时数组,用于保存邻接点的权值
            int[] weight = new int[graph.vertexNum];

            //临时数组,用于保存顶点信息
            int[] tempvertex = new int[graph.vertexNum];

            //取出邻接矩阵的第一行数据,也就是取出第一个顶点并将权和边信息保存于临时数据中
            for (int i = 1; i < graph.vertexNum; i++)
            {
                //保存于邻接点之间的权值
                weight[i] = graph.edges[0, i];

                //等于0则说明V1与该邻接点没有边
                if (weight[i] == short.MaxValue)
                    tempvertex[i] = noadj;
                else
                    tempvertex[i] = int.Parse(graph.vertex[0]);
            }

            //从集合V中取出V1节点,只需要将此节点设置为已访问过,weight为0集合
            var index = tempvertex[0] = used;
            var min = weight[0] = short.MaxValue;

            //在V的邻接点中找权值最小的节点
            for (int i = 1; i < graph.vertexNum; i++)
            {
                index = i;
                min = short.MaxValue;

                for (int j = 1; j < graph.vertexNum; j++)
                {
                    //用于找出当前节点的邻接点中权值最小的未访问点
                    if (weight[j] < min && tempvertex[j] != 0)
                    {
                        min = weight[j];
                        index = j;
                    }
                }
                //累加权值
                sum += min;

                Console.Write("({0},{1})  ", tempvertex[index], graph.vertex[index]);

                //将取得的最小节点标识为已访问
                weight[index] = short.MaxValue;
                tempvertex[index] = 0;

                //从最新的节点出发,将此节点的weight比较赋值
                for (int j = 0; j < graph.vertexNum; j++)
                {
                    //已当前节点为出发点,重新选择最小边
                    if (graph.edges[index, j] < weight[j] && tempvertex[j] != used)
                    {
                        weight[j] = graph.edges[index, j];

                        //这里做的目的将较短的边覆盖点上一个节点的邻接点中的较长的边
                        tempvertex[j] = int.Parse(graph.vertex[index]);
                    }
                }
            }
        }
        #endregion

二: 最短路径

1.   概念

        求最短路径问题其实也是非常有实用价值的,映射到交通系统图中,就是求两个城市间的最短路径问题,还是看这张图,我们可以很容易的看出比如

     V1到图中各顶点的最短路径。

      ① V1  ->  V2              直达,     权为2。

      ② V1  ->  V3              直达        权为3。

      ③ V1->V5->V4           中转       权为3+2=5。

      ④ V1  ->  V5               直达      权为3。

 

2.  Dijkstra算法

      我们的学习需要站在巨人的肩膀上,那么对于现实中非常复杂的问题,我们肯定不能用肉眼看出来,而是根据一定的算法推导出来的。

  Dijkstra思想遵循 “走一步,看一步”的原则。

     第一步: 我们需要一个集合U,然后将V1放入U集合中,既然走了一步,我们就要看一步,就是比较一下V1的邻接点(V2,V3,V5),

                 发现(V1,V2)的权值最小,此时我们将V2放入U集合中,表示我们已经找到了V1到V2的最短路径。

     第二步:然后将V2做中间点,继续向前寻找权值最小的邻接点,发现只有V4可以连通,此时修改V4的权值为(V1,V2)+(V2,V4)=6。

                此时我们就要看一步,发现V1到(V3,V4,V5)中权值最小的是(V1,V5),此时将V5放入U集合中,表示我们已经找到了

                V1到V5的最短路径。

     第三步:然后将V5做中间点,继续向前寻找权值最小的邻接点,发现能连通的有V3,V4,当我们正想修该V3的权值时发现(V1,V3)的权值

                小于(V1->V5->V3),此时我们就不修改,将V3放入U集合中,最后我们找到了V1到V3的最短路径。

     第四步:因为V5还没有走完,所以继续用V5做中间点,此时只能连通(V5,V4),当要修改权值的时候,发现原来的V4权值为(V1,V2)+(V2,V4),而

                现在的权值为5,小于先前的6,此时更改原先的权值变为5,将V4放入集合中,最后我们找到了V1到V4的最短路径。

复制代码 代码如下:

#region dijkstra求出最短路径
        /// <summary>
/// dijkstra求出最短路径
/// </summary>
/// <param name="g"></param>
        public void Dijkstra(MatrixGraph g)
        {
            int[] weight = new int[g.vertexNum];

            int[] path = new int[g.vertexNum];

            int[] tempvertex = new int[g.vertexNum];

            Console.WriteLine("/n请输入源点的编号:");

            //让用户输入要遍历的起始点
            int vertex = int.Parse(Console.ReadLine()) - 1;

            for (int i = 0; i < g.vertexNum; i++)
            {
                //初始赋权值
                weight[i] = g.edges[vertex, i];

                if (weight[i] < short.MaxValue && weight[i] > 0)
                    path[i] = vertex;

                tempvertex[i] = 0;
            }

            tempvertex[vertex] = 1;
            weight[vertex] = 0;

            for (int i = 0; i < g.vertexNum; i++)
            {
                int min = short.MaxValue;

                int index = vertex;

                for (int j = 0; j < g.vertexNum; j++)
                {
                    //顶点的权值中找出最小的
                    if (tempvertex[j] == 0 && weight[j] < min)
                    {
                        min = weight[j];
                        index = j;
                    }
                }

                tempvertex[index] = 1;

                //以当前的index作为中间点,找出最小的权值
                for (int j = 0; j < g.vertexNum; j++)
                {
                    if (tempvertex[j] == 0 && weight[index] + g.edges[index, j] < weight[j])
                    {
                        weight[j] = weight[index] + g.edges[index, j];
                        path[j] = index;
                    }
                }
            }

            Console.WriteLine("/n顶点{0}到各顶点的最短路径为:(终点 < 源点) " + g.vertex[vertex]);

            //最后输出
            for (int i = 0; i < g.vertexNum; i++)
            {
                if (tempvertex[i] == 1)
                {
                    var index = i;

                    while (index != vertex)
                    {
                        var j = index;
                        Console.Write("{0} < ", g.vertex[index]);
                        index = path[index];
                    }
                    Console.WriteLine("{0}/n", g.vertex[index]);
                }
                else
                {
                    Console.WriteLine("{0} <- {1}: 无路径/n", g.vertex[i], g.vertex[vertex]);
                }
            }
        }
        #endregion

最后上一下总的运行代码

复制代码 代码如下:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace MatrixGraph
{
    public class Program
    {
        static void Main(string[] args)
        {
            MatrixGraphManager manager = new MatrixGraphManager();

            //创建图
            MatrixGraph graph = manager.CreateMatrixGraph();

            manager.OutMatrix(graph);

            int sum = 0;

            manager.Prim(graph, out sum);

            Console.WriteLine("/n最小生成树的权值为:" + sum);

            manager.Dijkstra(graph);

            //Console.Write("广度递归:/t");

//manager.BFSTraverse(graph);

//Console.Write("/n深度递归:/t");

//manager.DFSTraverse(graph);

            Console.ReadLine();

        }
    }

    #region 邻接矩阵的结构图
    /// <summary>
/// 邻接矩阵的结构图
/// </summary>
    public class MatrixGraph
    {
        //保存顶点信息
        public string[] vertex;

        //保存边信息
        public int[,] edges;

        //深搜和广搜的遍历标志
        public bool[] isTrav;

        //顶点数量
        public int vertexNum;

        //边数量
        public int edgeNum;

        //图类型
        public int graphType;

        /// <summary>
/// 存储容量的初始化
/// </summary>
/// <param name="vertexNum"></param>
/// <param name="edgeNum"></param>
/// <param name="graphType"></param>
        public MatrixGraph(int vertexNum, int edgeNum, int graphType)
        {
            this.vertexNum = vertexNum;
            this.edgeNum = edgeNum;
            this.graphType = graphType;

            vertex = new string[vertexNum];
            edges = new int[vertexNum, vertexNum];
            isTrav = new bool[vertexNum];
        }

    }
    #endregion

    /// <summary>
/// 图的操作类
/// </summary>
    public class MatrixGraphManager
    {
        #region 图的创建
        /// <summary>
/// 图的创建
/// </summary>
/// <param name="g"></param>
        public MatrixGraph CreateMatrixGraph()
        {
            Console.WriteLine("请输入创建图的顶点个数,边个数,是否为无向图(0,1来表示),已逗号隔开。");

            var initData = Console.ReadLine().Split(',').Select(i => int.Parse(i)).ToList();

            MatrixGraph graph = new MatrixGraph(initData[0], initData[1], initData[2]);

            //我们默认“正无穷大为没有边”
            for (int i = 0; i < graph.vertexNum; i++)
            {
                for (int j = 0; j < graph.vertexNum; j++)
                {
                    graph.edges[i, j] = short.MaxValue;
                }
            }

            Console.WriteLine("请输入各顶点信息:");

            for (int i = 0; i < graph.vertexNum; i++)
            {
                Console.Write("/n第" + (i + 1) + "个顶点为:");

                var single = Console.ReadLine();

                //顶点信息加入集合中
                graph.vertex[i] = single;
            }

            Console.WriteLine("/n请输入构成两个顶点的边和权值,以逗号隔开。/n");

            for (int i = 0; i < graph.edgeNum; i++)
            {
                Console.Write("第" + (i + 1) + "条边:/t");

                initData = Console.ReadLine().Split(',').Select(j => int.Parse(j)).ToList();

                int start = initData[0];
                int end = initData[1];
                int weight = initData[2];

                //给矩阵指定坐标位置赋值
                graph.edges[start - 1, end - 1] = weight;

                //如果是无向图,则数据呈“二,四”象限对称
                if (graph.graphType == 1)
                {
                    graph.edges[end - 1, start - 1] = weight;
                }
            }

            return graph;
        }
        #endregion

        #region 输出矩阵数据
        /// <summary>
/// 输出矩阵数据
/// </summary>
/// <param name="graph"></param>
        public void OutMatrix(MatrixGraph graph)
        {
            for (int i = 0; i < graph.vertexNum; i++)
            {
                for (int j = 0; j < graph.vertexNum; j++)
                {
                    if (graph.edges[i, j] == short.MaxValue)
                        Console.Write("∽/t");
                    else
                        Console.Write(graph.edges[i, j] + "/t");
                }
                //换行
                Console.WriteLine();
            }
        }
        #endregion

        #region 广度优先
        /// <summary>
/// 广度优先
/// </summary>
/// <param name="graph"></param>
        public void BFSTraverse(MatrixGraph graph)
        {
            //访问标记默认初始化
            for (int i = 0; i < graph.vertexNum; i++)
            {
                graph.isTrav[i] = false;
            }

            //遍历每个顶点
            for (int i = 0; i < graph.vertexNum; i++)
            {
                //广度遍历未访问过的顶点
                if (!graph.isTrav[i])
                {
                    BFSM(ref graph, i);
                }
            }
        }

        /// <summary>
/// 广度遍历具体算法
/// </summary>
/// <param name="graph"></param>
        public void BFSM(ref MatrixGraph graph, int vertex)
        {
            //这里就用系统的队列
            Queue<int> queue = new Queue<int>();

            //先把顶点入队
            queue.Enqueue(vertex);

            //标记此顶点已经被访问
            graph.isTrav[vertex] = true;

            //输出顶点
            Console.Write(" ->" + graph.vertex[vertex]);

            //广度遍历顶点的邻接点
            while (queue.Count != 0)
            {
                var temp = queue.Dequeue();

                //遍历矩阵的横坐标
                for (int i = 0; i < graph.vertexNum; i++)
                {
                    if (!graph.isTrav[i] && graph.edges[temp, i] != 0)
                    {
                        graph.isTrav[i] = true;

                        queue.Enqueue(i);

                        //输出未被访问的顶点
                        Console.Write(" ->" + graph.vertex[i]);
                    }
                }
            }
        }
        #endregion

        #region 深度优先
        /// <summary>
/// 深度优先
/// </summary>
/// <param name="graph"></param>
        public void DFSTraverse(MatrixGraph graph)
        {
            //访问标记默认初始化
            for (int i = 0; i < graph.vertexNum; i++)
            {
                graph.isTrav[i] = false;
            }

            //遍历每个顶点
            for (int i = 0; i < graph.vertexNum; i++)
            {
                //广度遍历未访问过的顶点
                if (!graph.isTrav[i])
                {
                    DFSM(ref graph, i);
                }
            }
        }

        #region 深度递归的具体算法
        /// <summary>
/// 深度递归的具体算法
/// </summary>
/// <param name="graph"></param>
/// <param name="vertex"></param>
        public void DFSM(ref MatrixGraph graph, int vertex)
        {
            Console.Write("->" + graph.vertex[vertex]);

            //标记为已访问
            graph.isTrav[vertex] = true;

            //要遍历的六个点
            for (int i = 0; i < graph.vertexNum; i++)
            {
                if (graph.isTrav[i] == false && graph.edges[vertex, i] != 0)
                {
                    //深度递归
                    DFSM(ref graph, i);
                }
            }
        }
        #endregion
        #endregion

        #region prim算法获取最小生成树
        /// <summary>
/// prim算法获取最小生成树
/// </summary>
/// <param name="graph"></param>
        public void Prim(MatrixGraph graph, out int sum)
        {
            //已访问过的标志
            int used = 0;

            //非邻接顶点标志
            int noadj = -1;

            //定义一个输出总权值的变量
            sum = 0;

            //临时数组,用于保存邻接点的权值
            int[] weight = new int[graph.vertexNum];

            //临时数组,用于保存顶点信息
            int[] tempvertex = new int[graph.vertexNum];

            //取出邻接矩阵的第一行数据,也就是取出第一个顶点并将权和边信息保存于临时数据中
            for (int i = 1; i < graph.vertexNum; i++)
            {
                //保存于邻接点之间的权值
                weight[i] = graph.edges[0, i];

                //等于0则说明V1与该邻接点没有边
                if (weight[i] == short.MaxValue)
                    tempvertex[i] = noadj;
                else
                    tempvertex[i] = int.Parse(graph.vertex[0]);
            }

            //从集合V中取出V1节点,只需要将此节点设置为已访问过,weight为0集合
            var index = tempvertex[0] = used;
            var min = weight[0] = short.MaxValue;

            //在V的邻接点中找权值最小的节点
            for (int i = 1; i < graph.vertexNum; i++)
            {
                index = i;
                min = short.MaxValue;

                for (int j = 1; j < graph.vertexNum; j++)
                {
                    //用于找出当前节点的邻接点中权值最小的未访问点
                    if (weight[j] < min && tempvertex[j] != 0)
                    {
                        min = weight[j];
                        index = j;
                    }
                }
                //累加权值
                sum += min;

                Console.Write("({0},{1})  ", tempvertex[index], graph.vertex[index]);

                //将取得的最小节点标识为已访问
                weight[index] = short.MaxValue;
                tempvertex[index] = 0;

                //从最新的节点出发,将此节点的weight比较赋值
                for (int j = 0; j < graph.vertexNum; j++)
                {
                    //已当前节点为出发点,重新选择最小边
                    if (graph.edges[index, j] < weight[j] && tempvertex[j] != used)
                    {
                        weight[j] = graph.edges[index, j];

                        //这里做的目的将较短的边覆盖点上一个节点的邻接点中的较长的边
                        tempvertex[j] = int.Parse(graph.vertex[index]);
                    }
                }
            }
        }
        #endregion

        #region dijkstra求出最短路径
        /// <summary>
/// dijkstra求出最短路径
/// </summary>
/// <param name="g"></param>
        public void Dijkstra(MatrixGraph g)
        {
            int[] weight = new int[g.vertexNum];

            int[] path = new int[g.vertexNum];

            int[] tempvertex = new int[g.vertexNum];

            Console.WriteLine("/n请输入源点的编号:");

            //让用户输入要遍历的起始点
            int vertex = int.Parse(Console.ReadLine()) - 1;

            for (int i = 0; i < g.vertexNum; i++)
            {
                //初始赋权值
                weight[i] = g.edges[vertex, i];

                if (weight[i] < short.MaxValue && weight[i] > 0)
                    path[i] = vertex;

                tempvertex[i] = 0;
            }

            tempvertex[vertex] = 1;
            weight[vertex] = 0;

            for (int i = 0; i < g.vertexNum; i++)
            {
                int min = short.MaxValue;

                int index = vertex;

                for (int j = 0; j < g.vertexNum; j++)
                {
                    //顶点的权值中找出最小的
                    if (tempvertex[j] == 0 && weight[j] < min)
                    {
                        min = weight[j];
                        index = j;
                    }
                }

                tempvertex[index] = 1;

                //以当前的index作为中间点,找出最小的权值
                for (int j = 0; j < g.vertexNum; j++)
                {
                    if (tempvertex[j] == 0 && weight[index] + g.edges[index, j] < weight[j])
                    {
                        weight[j] = weight[index] + g.edges[index, j];
                        path[j] = index;
                    }
                }
            }

            Console.WriteLine("/n顶点{0}到各顶点的最短路径为:(终点 < 源点) " + g.vertex[vertex]);

            //最后输出
            for (int i = 0; i < g.vertexNum; i++)
            {
                if (tempvertex[i] == 1)
                {
                    var index = i;

                    while (index != vertex)
                    {
                        var j = index;
                        Console.Write("{0} < ", g.vertex[index]);
                        index = path[index];
                    }
                    Console.WriteLine("{0}/n", g.vertex[index]);
                }
                else
                {
                    Console.WriteLine("{0} <- {1}: 无路径/n", g.vertex[i], g.vertex[vertex]);
                }
            }
        }
        #endregion
    }
}

 

算法速成系列至此就全部结束了,公司给我们的算法培训也于上周五结束,呵呵,赶一下同步。最后希望大家能对算法重视起来,

学好算法,终身收益。

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