HDU 1081 To The Max

To The Max

Time Limit: 2000/1000 MS (Java/Others)    

Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6981    Accepted Submission(s): 3364

Problem Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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

15

 

Source

Greater New York 2001

 

Recommend

We have carefully selected several similar problems for you:  1024 1025 1080 1078 1074 

#pragma comment(linker,"/STACK:102400000,102400000")
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x7fffffff
#define mod 1000000007
#define ll long long
#define maxn 105
#define pi acos(-1.0)                          
int n, m, ans,s;
int a[maxn][maxn], sum[maxn],dp[maxn],c[maxn];
int main(){
    while (~scanf("%d", &n) && n){
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)scanf("%d", &a[i][j]);
        ans = 0;
        for (int i = 1; i <= n; i++){
            memset(c, 0, sizeof c);
            for (int j = i; j <= n; j++){
                dp[0] = 0; 
                for (int k = 1; k <= n; k++){
                    c[k]+=a[j][k];
                    dp[k] = max(dp[k - 1] + c[k], c[k]);
                    ans = max(ans, dp[k]);
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}