C-code for spoj problem named Martian Mining

martian.c

#include<stdio.h>

int Y[500][500];
int B[500][500];

int max(int a,int b){
  if(a>b)
    return a;
   return b;
}
int fun(int r,int c,int n,int m){
   if(r==n||c==m)
       return 0;   
    int i;
    int a = 0;
    for(i=r;i<n;i++)
       a += Y[i][c];
    int b = 0;
    for(i=c;i<m;i++)
      b += B[r][i];
   return max(a + fun(r,c+1,n,m),b + fun(r+1,c,n,m));

}
int fun2(int n,int m){

   int dp[n+1][m+1];

  int i,j,r,c;
  for(i=0;i<=n;i++)
    dp[i][m] = 0;
  for(j=0;j<=m;j++)
   dp[n][j] = 0;
  int a,b;
   for(r=n-1;r>=0;r--){
     for(c=m-1;c>=0;c--){
        for(i=r,a=0;i<n;i++)
             a += Y[i][c];
        for(i=c,b=0;i<m;i++)
            b += B[r][i];
        dp[r][c] = max(a + dp[r][c+1],b + dp[r+1][c]);
    }
 }
 return dp[0][0];
}

int main(){

int n,m;
  while(1){
    scanf("%d%d",&n,&m);
    if(n==0)
      break;
   int i,j;
  for(i=0;i<n;i++){
    for(j=0;j<m;j++){
        scanf("%d",&Y[i][j]);
    }
 }
  for(i=0;i<n;i++){
    for(j=0;j<m;j++){
        scanf("%d",&B[i][j]);
    }
  }
  printf("%d\n",fun2(n,m));
  }
  return 0;
}