Matrix Land


Problem Statement :


You are playing a matrix-based game with the following setup and rules:

You are given a matrix A with n rows and m columns. Each cell contains some points. When a player passes a cell their score increases by the number written in that cell and the number in the cell becomes 0. (If the cell number is positive their score increases, otherwise it decreases.)
The player starts from any cell in the first row and can move left, right or down.
The game is over when the player reaches the last row and stops moving.
image
Print the maximum score that the player can get.

Input Format

The first line contains n and m. The next n lines contain m numbers each, jth number in ith line denotes the number that is written on cell Aij.

Constraints

1 <= n*m <= 4*10^6
-250 <= Aij <= 250



Solution :



title-img


                            Solution in C :

In C++ : 





#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstdlib> 
#include<vector>
#include<map>
#include<cstring>
#define rep(i,j,k) for(register int i = j; i <= k; ++i)
#define dow(i,j,k) for(register int i = j; i >= k; --i)
#define ll long long
#define mp make_pair
#define fi first
#define se second 
#define all(x) x.begin(),x.end()
#define pb push_back
using namespace std;
 
inline int read() { 
    int s = 0, t = 1; char c = getchar();
    while( !isdigit(c) ) { if( c == '-' ) t = -1; c = getchar(); }
    while( isdigit(c) ) s = s * 10 + c - 48, c = getchar();
    return s * t;
}

const int N = 4e6+5, inf = 1e9+7;
int n, m, maxl, now, pre, f[2][N], v[N], g[N], h[N], sum[N]; 

int main() {
	n = read(), m = read(), now = 0, pre = 1;
	rep(i,1,n) { 
		swap(now,pre); 
		rep(j,1,m) v[j] = read();
		rep(j,1,m) sum[j] = sum[j-1] + v[j];
		rep(j,1,m) g[j] = max(g[j-1]+v[j],0);
		dow(j,m,1) h[j] = max(h[j+1]+v[j],0);
		maxl = -inf;
		rep(j,1,m) { 
			maxl = max(maxl,f[pre][j]-sum[j-1]+g[j-1]);
			f[now][j] = maxl+sum[j]+h[j+1];
		} 
		maxl = -inf;
		dow(j,m,1) { 
			maxl = max(maxl,f[pre][j]+sum[j]+h[j+1]);
			f[now][j] = max(f[now][j],maxl-sum[j-1]+g[j-1]);
		}
	}
	int ans = 0;
	rep(i,1,m) ans = max(ans,f[now][i]);
	cout<<ans<<endl;
	return 0;
}








In Java :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int m = in.nextInt();
        int[] a = new int[m];
        long[] left = new long[m];
        long[] right = new long[m];
        long[] prev = new long[m];
        long[] current = new long[m];
        long[] leftplus = new long[m];
        long[] rightplus = new long[m];
                
        for(int A_i = 0; A_i < n; A_i++){
            for(int A_j = 0; A_j < m; A_j++){
                a[A_j] = in.nextInt();
            }
            
            for (int i = 1; i < m; i++)
            {
                left[i] = Math.max(left[i - 1] + a[i - 1], 0);
            }
            
            for (int i = m - 2; i >= 0; i--)
            {
                right[i] = Math.max(right[i + 1] + a[i + 1], 0);
            }
            
            leftplus[0] = prev[0] + a[0];
            
            for (int i = 1; i < m; i++)
            {
                leftplus[i] = Math.max(prev[i] + a[i] + left[i], leftplus[i - 1] + a[i]);
            }
            
            rightplus[m - 1] = prev[m - 1] + a[m - 1];
            
            for (int i = m - 2; i >= 0; i--)
            {
                rightplus[i] = Math.max(prev[i] + a[i] + right[i], rightplus[i + 1] + a[i]);
            }
            
            for (int i = 0; i < m; i++)
            {
                current[i] = Math.max(leftplus[i] + right[i], rightplus[i] + left[i]);
            }
            
            long[] temp = current;
            current = prev;
            prev = temp;
        }
        
        long result = Long.MIN_VALUE;
        
        for (int i = 0; i < m; i++)
        {
            result = Math.max(prev[i], result);
        }
        
        System.out.println(result);
        in.close();
    }
}








In C :





#include <stdio.h>
#include <stdlib.h>
int max(int x,int y);
int N,*table[4000000],*dp,*tdp,*left_sum,*right_sum,*dp_left_tree;

int main(){
  int n,m,ma,total,i,j;
  scanf("%d%d",&n,&m);
  dp=(int*)malloc(m*sizeof(int));
  tdp=(int*)malloc(m*sizeof(int));
  left_sum=(int*)malloc(m*sizeof(int));
  right_sum=(int*)malloc(m*sizeof(int));
  dp_left_tree=(int*)malloc(m*sizeof(int));
  for(i=0;i<n;i++)
    table[i]=(int*)malloc(m*sizeof(int));
  for(i=0;i<n;i++)
    for(j=0;j<m;j++)
      scanf("%d",&table[i][j]);
  for(i=0;i<n;i++){
    for(j=total=0;j<m;j++){
      if(j)
        left_sum[j]=table[i][j]+left_sum[j-1];
      else
        left_sum[j]=table[i][j];
      total+=table[i][j];
    }
    for(j=m-1;j>=0;j--)
      if(j!=m-1)
        right_sum[j]=table[i][j]+right_sum[j+1];
      else
        right_sum[j]=table[i][j];
    for(j=m-2;j>=0;j--)
      left_sum[j]=max(left_sum[j],left_sum[j+1]);
    for(j=1;j<m;j++)
      right_sum[j]=max(right_sum[j],right_sum[j-1]);
    if(i){
      for(j=0;j<m;j++)
        dp_left_tree[j]=dp[j]+left_sum[j];
      for(j=m-2;j>=0;j--)
        dp_left_tree[j]=max(dp_left_tree[j],dp_left_tree[j+1]);
      for(j=0;j<m;j++)
        tdp[j]=right_sum[j]+dp_left_tree[j]-total;
      for(j=0;j<m;j++)
        dp_left_tree[j]=dp[j]+right_sum[j];
      for(j=1;j<m;j++)
        dp_left_tree[j]=max(dp_left_tree[j],dp_left_tree[j-1]);
      for(j=0;j<m;j++){
        if(left_sum[j]+dp_left_tree[j]-total>tdp[j])
          tdp[j]=left_sum[j]+dp_left_tree[j]-total;
      }
      for(j=0;j<m;j++)
        dp[j]=tdp[j];
    }
    else
      for(j=0;j<m;j++)
        dp[j]=left_sum[j]+right_sum[j]-total;
  }
  for(i=0,ma=-1000000001;i<m;i++)
    if(dp[i]>ma)
      ma=dp[i];
  printf("%d",ma);
  return 0;
}
int max(int x,int y){
  return (x>y)?x:y;
}
                        








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