Dijkstra算法求最短路径

Dijkstra算法求最短路径(C#版)行如下图的路径,(V0是中心):经过该算法后转化为下图using System;using System.Collections;using System.Text;namespace Greedy{class Marx{private int[] distance;private int row;private ArrayList ways = new ArrayList();public Marx(int n,params int[] d){this.row = n;distance = new int[row * row];for (int i = 0; i < row * row; i++){this.distance[i] = d[i];}for (int i = 0; i < this.row; i++) //有row 个点，则从中心到各点的路有row-1条{ArrayList w = new ArrayList();int j = 0;w.Add(j);ways.Add(w);}}//------------------------------public void Find_way(){ArrayList S = new ArrayList(1);ArrayList Sr = new ArrayList(1);int []Indexof_distance=new int[this.row];for(int i=0; i < row; i++){Indexof_distance[i]=i;}S.Add( Indexof_distance[0] );for (int i = 0; i < this.row; i++){Sr.Add( Indexof_distance[i] );}Sr.RemoveAt(0);int[] D = new int[this.row]; //存放中心点到每个点的距离//---------------以上已经初始化了，S和Sr(里边放的都是点的编号)------------------int Count = this.row - 1;while (Count>0){//假定中心点的编号是0的贪吃法求路径for (int i = 0; i < row; i++)D[i] = this.distance[i];int min_num = (int)Sr[0]; //距中心点的最小距离点编号foreach (int s in Sr){if (D[s] < D[min_num]) min_num = s;}//以上可以排序优化S.Add(min_num);Sr.Remove(min_num);//-----------把最新包含进来的点也加到路径中-------------((ArrayList)ways[min_num]).Add(min_num );//-----------------------------------------------foreach (int element in Sr){int position = element * (this.row) + min_num;bool exchange = false; //有交换标志if (D[element] < D[min_num] + this.distance[position])D[element] = D[element];else{D[element] = this.distance[position] + D[min_num];exchange = true;}//修改距离矩阵this.distance[element] = D[element];position = element * this.row;this.distance[position] = D[element];//修改路径---------------if (exchange == true){((ArrayList)ways[eleme nt]).Clear();foreach (int point in (ArrayList)ways[min_num])((ArrayList)wa ys[element]).Add(point);}}--Count;}}//----------------------------------------------------public void Display(){//------中心到各点的最短路径----------Console.WriteLine("中心到各点的最短路径如下: \n\n");int sum_d_index = 0;foreach(ArrayList mother in ways){foreach (int child in mother)Console.Write("V{0} -- ", child+1);Console.WriteLine(" 路径长{0}",distance[sum_d_index++]);}}}class MainEnterPoint{static void Main(string[] args){int r; //列数Console.Write("请输入点个数（含配送中心点）: ");Int32.TryParse(Console.ReadLine(), out r);Console.WriteLine("各点分别为: \n");for (int i = 0; i < r; i++)Console.Write("V{0} ", i);Console.Write(" 假定第一个点是配送中心");Console.WriteLine("\n\n输入各点之间的距离(无通径的用个大整数表示)\n");int[] a = new int[r * r];int da;for (int i = 0; i < r; i++){for (int j = i + 1; j < r; j++){Console.Write("V{0} 到V{1}的距离是: ",i,j);Int32.TryParse(Console.ReadLin e(), out da);a[i * r + j] = da;Console.WriteLine();}}//----完善距离矩阵(距离矩阵其实可以是个上三角矩阵，//----但为了处理方便，还是将其完整成一个对称阵)-----------for (int i = 0; i < r; i++){for (int j = 0; j < r; j++){if (i == j){a[i * r + j] = 0;}a[j * r + i] = a[i * r + j];}}Marx m=new Marx(r,a);Console.WriteLine();m.Find_way();m.Display();}}}//该程序不但能够算出从中心到各点的最短路径距离,而且把路径也保存了下来.dijkstra最短路径问题# include <stdio.h># include <stdlib.h># define maxlen 10# define large 999typedef struct{int vexnum;char vexs[maxlen];int arcs[maxlen][maxlen];}graph;void init_graph(graph *g){int i=0,j=0;g->vexnum=5;for(i=0;i<5;i++)for(j=0;j<5;j++)g->arcs[i][j]=1000;g->arcs[0][1]=1;g->arcs[0][3]=3;g->arcs[0][4]=7;g->arcs[1][2]=2;g->arcs[2][3]=2;g->arcs[2][4]=2;g->arcs[3][4]=5;g->arcs[3][2]=3;g->arcs[3][1]=1;g->vexs[0]='a';g->vexs[1]='b';g->vexs[2]='c';g->vexs[3]='d';g->vexs[4]='e';}void shortpath_dijkstra(graph g){int cost[maxlen][maxlen];//cost[i][j]: The cost of i to j.int dist[maxlen];//dist[i]: The distance of source point to i. int path[maxlen];//The point passed by.int s[maxlen];//if s[i]=1,then i is in the source point gather. int i,j,n,v0,min,u;printf("Input the source point(1 means the first point):"); scanf("%d",&v0);v0--;for(i=0;i<g.vexnum;i++){for(j=0;j<g.vexnum;j++)cost[i][j]=g.arcs[i][j];}for(i=0;i<g.vexnum;i++){dist[i]=cost[v0][i];if(dist[i]<large&&dist[i]>0) path[i]=v0;s[i]=0;}s[v0]=1;for(i=0;i<g.vexnum;i++){min=large;u=v0;for(j=0;j<g.vexnum;j++)if(s[j]==0&&dist[j]<min){min=dist[j];u=j;}s[u]=1;for(j=0;j<g.vexnum;j++)if(s[j]==0&&dist[u]+cost[u][j]<dist[j]){dist[j]=dist[u]+cost[u][j];path[j]=u;}}printf("Output\n",v0);for(i=0;i<g.vexnum;i++)if(s[i]==1){u=i;while(u!=v0){printf("%c<-",g.vexs[u]);u=path[u];}printf("%c",g.vexs[u]);printf(":%d\n",dist[i]);}else printf("%c<-%c:no path\n",g.vexs[i],g.vexs[v0]); }int main(){graph g;init_graph(&g);shortpath_dijkstra(g);}dijkstra最短路径问题program dijkstra;constinp = 'input.txt';oup = 'output.txt';maxn= 100;varga : array[1..maxn,1..maxn] of integer;dist : array[1..maxn] of integer;s : array[1..maxn] of 0..1;n,k : integer;fp : text;procedure init;vari,j: integer;beginassign(fp,inp); reset(fp);readln(fp,n,k);for i:=1 to n dofor j:=1 to n doread(fp,ga[i,j]);close(fp);end;procedure main;vari,j,w,m:integer;beginfillchar(s,sizeof(s),0);for i:=1 to n dodist[i]:=maxint;dist[k]:=0;for i:=1 to n-1 dobeginm:=maxint;for j:=1 to n doif (s[j]=0) and (dist[j]<m) thenbeginm:=dist[j];w:=j;end;s[w]:=1;for j:=1 to n doif (s[j]=0) and (ga[w,j]>0) and (dist[w]+ga[w,j]<dist[j]) then dist[j]:=dist[w]+ga[w,j];end;end;procedure print;vari,j:integer;beginassign(fp,oup);rewrite(fp);for i:=1 to n dowrite(fp,dist[i],' ');close(fp);end;begininit;main;print;end.。