最短路径问题

来源：互联网发布：思来氏知乎编辑：程序博客网时间：2024/05/30 04:18

#include<stdio.h>#include<malloc.h>#include<string.h>#include<stack>using namespace std;#define MAX_VERTEX_NUM 20#define INFINITY 32768int visited[100];int path[MAX_VERTEX_NUM];typedef struct node1{int adj;}gra;typedef struct node2{gra arcs[100][100];int vertex[100];int vexnum,arcnum;}*graph,graph1;int locatevertex(graph &g,int v){int j=0,k;for(k=0;k<g->vexnum;k++){if(g->vertex[k]==v){j=k;break;}}return j;}void create(graph &g){int i,j,k,v1,v2,weight;printf("请输入图的最大顶点数和最大弧数: ");scanf("%d%d",&g->vexnum,&g->arcnum);for(i=0;i<g->vexnum;i++){for(j=0;j<g->vexnum;j++){if(i==j)g->arcs[i][j].adj=0;elseg->arcs[i][j].adj=INFINITY;}}printf("请输入图的各顶点值: ");for(i=0;i<g->vexnum;i++)scanf("%d",&g->vertex[i]);for(k=0;k<g->arcnum;k++){printf("请输入两顶点1,2表示1到2有关系: ");scanf("%d%d%d",&v1,&v2,&weight);i=locatevertex(g,v1);j=locatevertex(g,v2);g->arcs[i][j].adj=weight;}}void shortestpath(graph &g,int n){int v0=0,k,min;int dist[MAX_VERTEX_NUM];    bool S[MAX_VERTEX_NUM];for(int i=0;i<g->vexnum;i++){S[i]=false;//对S初始化dist[i]=g->arcs[v0][i].adj;//对dist初始化if(dist[i]<INFINITY)path[i]=0;elsepath[i]=-1;}S[v0]=true;for(int t=1;t<=g->vexnum-1;t++){min=INFINITY;for(i=0;i<g->vexnum;i++){if(!S[i]&&dist[i]<min){ min=dist[i];             k=i;}}if(min==INFINITY)return ;for(i=0;i<g->vexnum;i++){if(!S[i]&&g->arcs[k][i].adj!=INFINITY&&(dist[k]+g->arcs[k][i].adj<dist[i])&&k!=i){dist[i]=dist[k]+g->arcs[k][i].adj;path[i]=k;}}S[k]=true;}n=locatevertex(g,n);int x=n;while(n>=0){printf("%d ",path[n]);if(path[n]==0)break;n=path[n];}printf("最短路径长度为: ");printf("%d\n",dist[x]);}int main(){graph g;int n;g=(graph)malloc(sizeof(graph1));create(g);printf("请输入一个顶点求出它的最短路径: ");scanf("%d",&n);shortestpath(g,n);return 0;}

算法详解：<p>首先S中有v1,则计算v1到V-S即v2,v3,v4,v5中距离，然后选取最小的距离，作为v1到v2,v3,v4,v5的一条最短路径即v1->v2,然后更新dist[i],i=3,4,5，即把v2到v3,v4,v5的距离</p><p>与v1到v3,v4,v5的距离作比较,即取较小距离作为作为更新的数据,然后从dist[i],i=3,4,5选取最小值作为v1,v2到v3,v4,v5的最短路径,注意v2表示从v1到v2.即选出v3,即v2->v3,</p><p>加入到S中,然后更新dist[i],i=4,5，即把v3到v4,v5的距离与v2到v3,v4,v5的距离作比较,即取较小距离,作为更新的数据,然后依次类推，并标记最短路径,直到V-S为空,即可求</p><p>出v1到所有顶点的最短距离</p>

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