UVa 10664 - Luggage

10664 - Luggage

 * http://acm.uva.es/p/v106/10664.html

Summary
There is a list of luggage (defined by their weights) and you are trying to split them among 2 cars exactly equally,

print "YES" if you can, "NO" otherwise

Explanation
This problem can be solving using dp,

You run a dp algo trying to get the maximum value <= 1/2 sum of all weights

if the maximum value is equal to 1/2 the sum then print "YES" otherwise print "NO"

int dp(int r, int c){ // r is rem, c is current element if(mem[r][c] != -1) return mem[r][c]; int mx = 0, t; for(int i = c; i < n; i++){ if(arr[i] <= r){ t = arr[i] + dp(r-arr[i], i+1); if(mx < t) mx = t;   } } mem[r][c] = mx; return mem[r][c]; } int main{ . . . if(sum % 2 == 1) cout << "NO" << endl; else if(dp(sum/2, 0) == sum/2) cout << "YES" << endl; else cout << "NO" << endl; . }

Input
3 1 2 1 2 1 2 3 4 1 2 5 10 50 3 50 3 5 2 7 1 7 5 2 8 9 1 25 15 8 3 1 38 45 8 1

Output
NO YES YES

Solution
Java : http://adilakhter.wordpress.com/2013/03/08/uva-10664-luggage-with-dynamic-programming-and-java/