Dijkstra shortest path and its output

High school students let me beg 8 market and 35 points of sale (they will also have 15 intersections) between the 35*8=280 a combination of the shortest path and its output.

The shortest water problem, hey. It's great to have the knowledge you have learned to help others solve problems.

Specific comments can be seen in the code, written pretty clear.

#include <iostream>#include<cstdio>#include<string>#include<cstring>#include<algorithm>using namespacestd;#defineINF 0x3f3f3f3//inf represents InfinityConst intmax_n= -;intMap[max_n][max_n],vis[max_n];intDis[max_n];//dis represents the shortest distance to each pointintPre[max_n];//Recording precursorintAns[max_n];//an array of output pathsvoidDIJKSTR (intXintm) {//x represents the start point, and M represents the number of vertices     for(intI=1; i<=m;i++) {Dis[i]=Map[x][i]; Vis[i]=0; if(map[x][i]!=INF) Pre[i]=x; ElsePre[i]=-1; }    intp; VIS[X]=1;  for(intI=1; i<=m;i++){           intmin=INF;  for(intj=1; j<=m;j++){               if(!vis[j]&&dis[j]<min) {min=Dis[j]; P=J; }} Vis[p]=1;  for(intj=1; j<=m;j++){            if(!vis[j]&&dis[p]+map[p][j]<Dis[j]) {Dis[j]=dis[p]+Map[p][j]; PRE[J]=p; }                    }               }}intMain () {intM,n,a,b,c; //The number in the input file means: 1-8 is base, 9-43 is point of sale, 44-58 is intersection, format is point distanceFreopen ("In.txt","R", stdin);//read data from a fileFreopen ("OUT.txt","W", stdout);//output results to file    voidDIJKSTR (intXintN); //cout<< "Please enter the number of points and the number of edges" <<endl;Cin>>m>>N; memset (Vis,0,sizeof(VIS));//m points, n edges         for(intI=1; i<=m;i++)         for(intj=1; j<=m;j++) {Map[i][j]=INF; }    //cout<< "Please enter the edges in order of origin, end, distance" <<endl;         for(intj=1; j<=n;j++) {cin>>a>>b>>C; MAP[A][B]=map[b][a]=C; }                for(intI=1; i<=8; i++) {dijkstr (i,m);  for(intj=9; j<= +; j + +){            intk=J; cout<<"from the base"<<i<<"to the point of sale"<< (J-8) <<"the shortest distance is:"<<dis[k]<<Endl; intp,len=1; P=K;  while(p>=1) {Ans[len++]=Q; P=Pre[p]; }       intfirst=0; cout<<"here is the exact path:"<<Endl;  for(intr=len-1; r>=1; r--)       {            if(first++) cout<<" -"; if(ans[r]>=1&&ans[r]<=8) cout<<"Base"<<Ans[r]; Else if(ans[r]>=9&&ans[r]<= +) cout<<"Point of Sale"<<ans[r]-8; Elsecout<<"intersection"<<ans[r]- +; } cout<<Endl; cout<<Endl; }    }        return 0;}

Preview of results in out file:

Detailed data and source code available for download: Http://files.cnblogs.com/files/Tach-ac/%E6%9C%80%E7%9F%AD%E8%B7%AF.rar

Dijkstra shortest path and its output