PAT gas Station

Gas station

A gas station have to being built at such a location that the minimum distance between the station and any of the residential Housing is as far away as possible. However it must guarantee that all the houses is in its service range.

Now given the map of the city and several candidate locations for the gas station, you is supposed to give the best recom Mendation. If there is more than one solution, the output of the one with the smallest average distance to all the houses. If Such a solution is still not unique, the output of the one with the smallest index number.

Test instructions a very complicated question. I sort of sorted it out. That's pretty much it.

1. Figure out the distance from each gas station to all house.

2. All distances must be less than the maximum service distance of the gas station

3. Find a minimum distance from all distances

4. Take the maximum distance at the minimum distance per gas station

5. If the minimum distance is equal, the average distance is minimized.

6. If the average distance is the smallest, take the smallest number

Are the steps above for this problem almost dizzy? The algorithm should still be simple is the Dijkstra algorithm

The AC code is given below

1#include <stdio.h>2#include <stdlib.h>3 #defineMAX 10000004 5 6 intg[1011][1011];7 intN,m,k,ds;8 intdis[1011];9 intflag[1011];Ten floatsum[Ten]; One  A voidDijkstraintv) - { -     inti,j,pos,min; thedis[v]=0; -      -      for(i=1; i<=n+m;i++) -     {     +min=MAX; -          for(j=1; j<=n+m;j++) +         { A             if(flag[j]!=1&& dis[j]<min) at             { -min=Dis[j]; -pos=J; -             } -         } -flag[pos]=1; in          for(j=1; j<=n+m;j++) -         { to             if(flag[j]!=1) +             { -                 if(dis[pos]+g[pos][j]<Dis[j]) the                 { *dis[j]=dis[pos]+G[pos][j]; $                 }Panax Notoginseng             } -         } the              +     } A } the intMain () + { -     inti,j; $     ints3,s4,l,k,pos=0, t=0, flag1=0; $     intmin[Ten]; -     floatAve,ave2,max; -     Chars1[Ten],s2[Ten]; the      -scanf"%d%d%d%d",&n,&m,&k,&Ds); Wuyi      for(i=1; i<=n+m;i++) the     { -dis[i]=MAX; Wu          for(j=1; j<=n+m;j++) -g[i][j]=MAX; About     } $      for(i=1; i<=k;i++) -     { -scanf"%s%s%d",s1,s2,&L); -          A         if(s1[0]=='G') +S3=atoi (s1+1)+N; the         Else -s3=atoi (S1); $         if(s2[0]=='G') theS4=atoi (s2+1)+N; the         Else thes4=atoi (S2);  theg[s3][s4]=g[s4][s3]=L; -     } in      the      for(i=n+1; i<=n+m;i++) the     { About Dijkstra (i); themin[t]=dis[1]; the          for(j=1; j<=n;j++) the         { +             if(dis[j]>Ds) -             { themin[t]=-1;Bayisum[t]=-1; the                      Break; the             }     -             if(min[t]>Dis[j]) -             { themin[t]=Dis[j];  the             } thesum[t]+=Dis[j];  the         } -t++; the          for(j=1; j<=n+m;j++) the         { thedis[j]=MAX; 94flag[j]=0; the         }         the     } the      for(i=0; i<m;i++)98     { About         if(min[i]!=-1) -         {101flag1=1;102              Break;103             104         } the     }106max=-100000;107      for(i=0; i<m;i++)108     {109         if(flag1==0) the              Break;111         if(max<min[i]&&min[i]!=-1) the         {113max=Min[i]; thepos=i; theave=sum[pos]/N; the         }    117         Else if(max==min[i]&&min[i]!=-1)118         {119ave=sum[pos]/N; -ave2=sum[i]/N;121             if(ave>ave2)122             {123max=Min[i];124pos=i; theAve=ave2;126             }    127                      -         }129     } the     if(flag1==1)131printf"g%d\n%.1f%.1f\n", pos+1, Max,ave);  the     Else133printf"No solution\n");134}

PAT gas Station