A Mobile Phone Controlled Robot

This is a robot which is able to pick an object carry it and drop it. It is remotely operated by cell phone. The idea of DTMF tone decoder is used here.

A Line Follower Robot

This is a line follower robot which follows a white line on a green background. We used LDR as sensor. ATMEGA 32 microcontroller as its brain. When it gets a black mark on the side of its path it takes the next turn on that side. The robot was prepare for GRC13 Bangladesh round.

UVA 116 Solution CPP code

Create two arrays in addition to the input array. One is the DP array, and one is the path array. dp[i][j] = in[i][j] + min(dp[i-1][j-1], dp[i][j-1], dp[i+1][j-1]). Make sure that you wrap around the top and bottom of the matrix (use modulo, or just an if statement).

For the path, you must keep an array of strings (or whatever you like) to hold the current path to that point. Prefer lexicographically shorter paths when two DP cells have the same value. Note that lexicographically shorter in this problem means that you minimize the row number, and not the ASCII value of the string. For example:

1 10 10 10
1 2 3 4

In ASCII terms, and normal string comparison terms, the first string is lexicographically earlier. But in this problem, they want the latter.

Remember it does not matter you from left to right or right to left.

CODE:

#include<stdio.h>


#define MIN(a,b) a<b? a:b
#define maxe 1<<30

int p[11][101],r,c;



int main(){


    int i,j,rp;
    long long a[11][101],tmp,min,f;

 while(scanf("%d %d",&r,&c)!=EOF){

     for(i=0;i<r;i++)
     for(j=0;j<c;j++)
     scanf("%lld",&a[i][j]);

    for(j=c-2;j>=0;j--)
    for(i=0;i<r;i++){

        tmp=a[i][j];

        rp=i-1;
        if(rp<0)rp=r-1;
        a[i][j]=tmp+a[rp][j+1];
        p[i][j]=rp;



        rp=i;f=1;
        if(a[i][j]>tmp+a[rp][j+1]||(a[i][j]==tmp+a[rp][j+1]&&i==0)){
        a[i][j]=tmp+a[rp][j+1];
        p[i][j]=rp;
        f=0;
        }


        rp=i+1;
        if(rp==r)rp=0;
        if(a[i][j]>tmp+a[rp][j+1]||(a[i][j]==tmp+a[rp][j+1]&&i==0&&f)||(a[i][j]==tmp+a[rp][j+1]&&i==r-1)){
        a[i][j]=tmp+a[rp][j+1];
        p[i][j]=rp;
        }



        }

     min=maxe;
     for(i=0;i<r;i++){
        if(min>a[i][0]){
        min=a[i][0];
        rp=i;
        }
     }

    for(j=0;j<c;j++){

        printf("%d",rp+1);
        if(j<c-1)printf(" ");
        rp=p[rp][j];
        }

    printf("\n%lld\n",min);


     }

}