LightOJ 10047 solution & CPP code

Its a dp problem.
Keep on counting minimum possible cost for each position. Use previous value to avoid calculating same thing again and again.

#include<stdio.h>
#define MIN(a,b) a<b? a:b
#define inf 99999999


int main()
{

    int fc,cost[22][5],a[22][5],i,j,n,t,kase,x,y;
    scanf("%d",&t);
    kase=0;
    while (kase++<t)
    {

        scanf("%d",&n);


        for (i=1;i<=n;i++)
            for (j=1;j<=3;j++)
                scanf("%d",&a[i][j]);


        cost[1][1]=a[1][1];
        cost[1][2]=a[1][2];
        cost[1][3]=a[1][3];

        for (i=2;i<=n;i++)
            {
                cost[i][1]=cost[i-1][2]+a[i][1];
                cost[i][1]=MIN(cost[i][1],cost[i-1][3]+a[i][1]);

                cost[i][2]=cost[i-1][1]+a[i][2];
                cost[i][2]=MIN(cost[i][2],cost[i-1][3]+a[i][2]);

                cost[i][3]=cost[i-1][2]+a[i][3];
                cost[i][3]=MIN(cost[i][3],cost[i-1][1]+a[i][3]);

            }
        fc=inf;
        for (i=1;i<=3;i++)
            fc=MIN(fc,cost[n][i]);

        printf("Case %d: %d\n",kase,fc);

    }

    return 0;
}