UVA 1366 Martian Mining (DP)

As long as this problem is selected i,j then either the horizontal, or the column is the i,j before the row all selected, cannot turn.

So each update corresponds to a length or width extension of 1 from an existing rectangle, then the state equation can be defined as Dp[i][j]=max (Dp[i-1][j]+row[i][j],dp[i][j-1]+col[i][j]).

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN=500+10;
#define REP (i,a,b) for (int i= (a), i<= (b); i++)
#define RET (I,A,B) for (int i= (a); i>= (b); i--)
#define SS (x) scanf ("%d", &x)
int ROW[MAXN][MAXN],COL[MAXN][MAXN];
int n,m;
int F[MAXN][MAXN];
int main ()
{
    while (true)
    {
        int x;
        SS (N); SS (M); if (n+m==0) break;
        Rep (I,1,n) Rep (j,1,m) {SS (x); row[i][j]=row[i][j-1]+x;}
        Rep (I,1,n) Rep (j,1,m) {SS (x); col[i][j]=col[i-1][j]+x;}
        Rep (i,1,n) {
            Rep (j,1,m) {
                F[i][j]=max (f[i][j-1]+col[i][j],f[i-1][j]+row[i][j]);
            }
        }
        cout<<f[n][m]<<endl;
    }
    return 0;
}