Find the Mincost route
Time limit:1000/2000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 2995 Accepted Submission (s): 1204
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 Input
3 31 2 12 3 11 3 13 31 2 11 2 32 3 1
Sample Output
3It ' s impossible.
Author8600
Sourcehdu 2007-spring Programming Contest-warm up (1)
The topic requires a ring, so there are some improvements to the Floyd algorithm. can refer to the explanation here: http://blog.sina.com.cn/s/blog_476a25110100mag6.html
#include <stdio.h>
#include <string.h>
int n,m,mat[105][105],dis[105][105];
int main ()
{
int a,b,c,i,j,k;
while (scanf ("%d%d", &n,&m)!=eof)
{
for (i=0;i<105;i++)
for (j=0;j<105;j++)
{mat[i][j]=99999999;
dis[i][j]=99999999;
}
for (int i=0;i<m;i++)
{
scanf ("%d%d%d", &a,&b,&c);
if (mat[a][b]>c)
Mat[a][b]=mat[b][a]=dis[b][a]=dis[a][b]=c;
}
int i,k,j,temp=99999999;
for (k=1;k<=n;k++)
{
for (i=1;i<k;i++)
for (j=1;j<i;j++)
{
if (Temp>dis[i][j]+mat[i][k]+mat[k][j])
TEMP=DIS[I][J]+MAT[I][K]+MAT[K][J]; The equivalent of a ring with the shortest path found at the time of the K-th.
}
for (i=1;i<=n;i++)
for (j=1;j<=n;j++)
if (Dis[i][j]>dis[i][k]+dis[k][j])//To update the shortest path
DIS[I][J]=DIS[I][K]+DIS[K][J];
}
if (temp<99999999)
printf ("%d\n", temp);
else printf ("It ' s impossible.\n");
}
return 0;
}
HDU 1599find the Mincost route (Floyd algorithm, least-loop graph)
