最短路径算发实现

#include <stdio.h>#include <conio.h>#include <stdlib.h>#define X 10000#define VertexNum 7 //实际上共有六个顶点（1---6）#define EdgeNum 9int Graph[VertexNum][VertexNum] =//0 1 2 3 4 5 6{ X, X, X, X, X, X, X, //0 X, X, 6, 3, X, X, X, //1 X, X, X, X, 5, X, X, //2 X, X, 2, X, 3, 4, X, //3 X, X, X, X, X, X, 3, //4 X, X, X, X, 2, X, 5, //5 X, X, X, X, X, X, X //6};int Visited[VertexNum];int path[VertexNum];int Distance[VertexNum];void Dijkstra(int Begin){ int MinEdge, Vertex, i,j, Edges; Edges = 1; Visited[Begin] = 1; for (i = 1; i<VertexNum; i++) Distance[i] = Graph[Begin][i]; Distance[Begin] = 0; printf(" 1 2 3 4 5 6//n"); printf("-----------------------------------//n"); printf("s:%d", Edges); for( i=1; i<VertexNum; i++) if (Distance[i] == X) printf(" *"); else printf("%3d",Distance[i]); printf("//n"); while( Edges<VertexNum-1) { Edges++; MinEdge = X; for(j=1; j<VertexNum; j++) if (Visited[j]==0 && MinEdge > Distance[j] ) { Vertex = j; MinEdge = Distance[j]; } Visited[Vertex] = 1; printf("s:%d",Edges); for(j=1; j<VertexNum; j++) { if (Visited[j] == 0 && Distance[Vertex] + Graph[Vertex][j] <Distance[j]) { Distance[j] = Distance[Vertex] + Graph[Vertex][j]; path[j] = Vertex; } //printf("%6d",Distance[j]); if (Distance[j] == X) printf(" *"); else printf("%3d",Distance[j]); } printf("//n"); }}void main(){ int i; int k; // clrscr(); for(i=0; i<VertexNum; i++) { Visited[i] = 0; path[i] = 1;} Dijkstra(1); printf("//n//nAll Path-------------------------//n"); for(i=2; i<VertexNum; i++) //printf("%5d",Visited[i]); { printf("[%d] ",Distance[i]); k = i; do { printf("%d<--",k); k = path[k]; } while (k!=1); printf("1 //n"); }}