c++

Floyd最短路径的算法实现C++#include<stdio.h>#defineINF10000inte[5][5]={0,45,30,INF,50,45,0,INF,60,INF,30,INF,0,INF,45,INF,60,INF,0,55,50,INF,45,55,0};intpath[5][5]={0};voidfloyd(){inti,j,k;for(i=0;i<5;i++)for(j=0;j<5;j++)for(k=0;k<5;k++)if(e[i][k]<INF&&e[k][j]<INF&&e[i][j]>e[i][k]+e[k][j]){e[i][j]=e[i][k]+e[k][j];path[i][j]=k;}}voidgetPath(inti,intj){if(i==j)return;if(path[i][j]==0)printf("%c",j+'a');else{getPath(i,path[i][j]);getPath(path[i][j],j);}}intmain(){charstart,end;printf("输入起点编号和终点编号：");scanf("%c%c",&start,&end);floyd();intm,n;for(m=0;m<4;m++){for(n=0;n<4;n++){printf("%d",path[m][n]);}}printf("最短路径为：%d，具体如下：\n%c",e[start-'a'][end-'a'],start);getPath(start-'a',end-'a');return0;}