Find the Mincost route (min ring, shortest way, Floyd)

Find the Mincost route

Time limit:1000/2000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 3425 Accepted Submission (s): 1397

Problem description Hangzhou has n scenic spots, there are some two-way connection between the scenic spots, now 8600 want to find a tourist route, this route from a point and finally back to a point, assuming the route is V1,v2,.... VK,V1, then must meet the k>2, that is, apart from the starting point to go through at least 2 other different scenic spots, and can not be repeated through the same scenic area. Now 8600 needs you to help him find a route like this, and the less it costs the better.

The first line of input is 2 integers n and m (n <=, M <= 1000), representing the number of scenic spots and the number of roads.
In the next M-line, each row consists of 3 integers a,b,c. Represents a path between A and B and costs C (c <= 100).

Output for each test instance, the minimum value is spent if such a route can be found. If it is not found, output "It's impossible."

Sample INPUT3 31 2 12 3 11 3 13 31 2 11 2 32 3 1

Sample output3it ' s impossible.

Author8600: Wrong to helpless ... In fact, this problem let the ball at least through the shortest ring of three cities, with the idea is to use a map to record the initial edge weights, and then use DIS record the shortest road, and then Anser=min (Anser,dis[i][j]+map[j][k]+map[k][i]); The minimum ring weight of three cities is guaranteed, and the dis "i" [j] represents the shortest circuit code from I to J City:

1#include <stdio.h>2#include <string.h>3 #defineMIN (x, y) (x<y?x:y)4 Const intinf=0x3f3f3f;5 Const intmaxn= the;6 intMap[maxn][maxn],dis[maxn][maxn],anser;7 intn,m;8 voidInitial () {9      for(intI=1; i<= -; i++)Ten          for(intj=1; j<= -; j + +) One             if(I-J) map[i][j]=INF; A                 Elsemap[i][j]=0; - } - voidFloyd () { the      for(intI=1; i<=n;i++) -          for(intj=1; j<=n;j++) -dis[i][j]=Map[i][j]; -Anser=INF; +      for(intk=1; k<=n;k++){ -          for(intI=1; i<=n;i++) +                  for(intj=1; j<=n;j++) A                     if(i!=j&&i!=k&&j!=k) atAnser=min (anser,dis[i][j]+map[j][k]+map[k][i]); -                     /*if (Anser>dis[i][j]+map[j][k]+map[k][i]) { - printf ("dis[%d][%d]=%d\n", I,j,dis[i][j]); - printf ("map[%d][%d]=%d\n", J,k,map[j][k]); - printf ("map[%d][%d]=%d\n", K,i,map[k][i]); -                     }*/ in          for(intI=1; i<=n;i++) -              for(intj=1; j<=n;j++) toDis[i][j]=min (Dis[i][j],dis[i][k]+dis[k][j]);//Make maple wrong half a day ....  +                     } -     if(Anser==inf) puts ("It ' s impossible."); the     Elseprintf"%d\n", Anser); * } $ intMain () {Panax Notoginseng     inta,b,c; -      while(~SCANF ("%d%d",&n,&M)) { the initial (); +          while(m--){ Ascanf"%d%d%d",&a,&b,&c); the             if(C<map[a][b]) map[a][b]=map[b][a]=C; +         } - Floyd (); $     } $     return 0; -}

Find the Mincost route (min ring, shortest way, Floyd)