POJ 1050 To the Max

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
As an example, the maximal sub-rectangle of the array:

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
is in the lower left corner:

9 2
-4 1
-1 8
and has a sum of 15.

Input

The input consists of an N * N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N^2 integers separated by whitespace (spaces and newlines). These are the N^2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].

Output

Output the sum of the maximal sub-rectangle.

Sample Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4  1 -1

8  0 -2

Sample Output

Question 15 means to find the maximum submatrix sum. Analyze the extension of one-dimensional maximal consecutive subsegment sums. First, when inputting the traversal, you can find the max (the sum of the largest consecutive sub-segments of each row, which is equivalent to a sub-matrix with a width of one), and then traverse, adding the elements of the corresponding columns of each row after the i-th row to the corresponding column of the i-th row Elements, each time a row is added, the maximum field sum is calculated once, so that the multiple rows of the sub-matrix are compressed into one row, and after it becomes one row, the maximum field sum is obtained! clever!

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include <queue>
#include <vector>
#include<bitset>
#include<map>
#include<deque>
using namespace std;
typedef long long LL;
const int maxn = 1e4+5;
const int mod = 77200211+233;
typedef pair<int,int> pii;
#define X first
#define Y second
#define pb push_back
//#define mp make_pair
#define ms(a,b) memset(a,b,sizeof(a))
const int inf = 0x3f3f3f3f;
#define lson l,m,2*rt
#define rson m+1,r,2*rt+1
typedef long long ll;
#define N 100010
int a[105][105];
int main(){
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif // LOCAL
    int n,tmp;
    scanf("%d",&n);
    int ans = -inf;
    for(int i=1;i<=n;i++){
        tmp=0;
        for(int j=1;j<=n;j++){
            scanf("%d",&a[i][j]);
            if(tmp>0) tmp+=a[i][j];
            else tmp=a[i][j];
            years = max(years,tmp);
        }

    }

    for(int i=1;i<n;i++){
        for(int j=i+1;j<=n;j++){
            tmp=0;
            for(int k=1;k<=n;k++){
                a[i][k]+=a[j][k];
                if(tmp>0) tmp+= a[i][k];
                else tmp = a[i][k];
                years = max(years,tmp);
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}

By the way, give the code and idea of finding the maximum sub-segment sum:

int maxsum ( int x [], int n)  
{  
   int i,b = 0,k = -10000000;  
   
   for(i = 0 ; i < n ; ++i)  
   {  
       if (b > 0 ) b += x[i]; // If the accumulated sum is positive, continue to add    
       /*  
          If b <= 0, then there must be x[i-1]<0, x[i] is pending, then if x[i]>= 0,
          b=x[i] for granted; what if x[i]<0? Is b=x[i] appropriate? The answer is appropriate.
          Because b is still less than 0 in the next loop, you can definitely find a number greater than 0     
            
          There is one more question: b = x[i], why don't you just discard all the fields just now?
          What if the sub-segments of the sub-segment just now (the first few are negative numbers) are greater than 0? But this is not possible.
          Because the first number of a field must be a positive number, because if the first number is negative,
          Then b<0, the else will be executed, and the accumulation of a sub-segment will not start until a positive number appears.  
       */  
           
       else   b = x[i]; // If the accumulated sum is negative, assign this value to b    
               
       if (b > k) k = b; // update max field sum    
   }  
          
   return k;   
}

Guess you like