OUTPUT:
enter the cost matrix :
0 1 4 2 0
0 0 0 2 3
0 0 0 3 0
0 0 0 0 5
0 0 0 0 0
enter number of paths : 4
enter possible paths :
1 2 4 5 0
1 2 5 0 0
1 4 5 0 0
1 3 4 5 0
minimum cost : 4
minimum cost path :
1–>2–>5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 | #include<stdio.h> #include<conio.h> void main() { int path[5][5],i,j,min,a[5][5],p,st=1,ed=5,stp,edp,t[5],index; clrscr(); printf("enter the cost matrix\n"); for(i=1;i<=5;i++) for(j=1;j<=5;j++) scanf("%d",&a[i][j]); printf("enter number of paths\n"); scanf("%d",&p); printf("enter possible paths\n"); for(i=1;i<=p;i++) for(j=1;j<=5;j++) scanf("%d",&path[i][j]); for(i=1;i<=p;i++) { t[i]=0; stp=st; for(j=1;j<=5;j++) { edp=path[i][j+1]; t[i]=t[i]+a[stp][edp]; if(edp==ed) break; else stp=edp; } } min=t[st];index=st; for(i=1;i<=p;i++) { if(min>t[i]) { min=t[i]; index=i; } } printf("minimum cost %d",min); printf("\n minimum cost path "); for(i=1;i<=5;i++) { printf("--> %d",path[index][i]); if(path[index][i]==ed) break; } getch(); } |
OUTPUT:
enter the cost matrix :
0 1 4 2 0
0 0 0 2 3
0 0 0 3 0
0 0 0 0 5
0 0 0 0 0
enter number of paths : 4
enter possible paths :
1 2 4 5 0
1 2 5 0 0
1 4 5 0 0
1 3 4 5 0
minimum cost : 4
minimum cost path :
1–>2–>5
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September 30th, 2009 at 4:55 pm
hi..i found this c code after a long time search…i am doing a project work in shortest path detection… i can’t understand this..can u much detail abt this…its very helpful to me….diagramatic representation of ur eg is much better.plz do this help as soon as possible…
January 18th, 2011 at 3:19 pm
very efficient code