【POJ 1050】To the Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 43725		Accepted: 23166

Description

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

15

Source

Greater New York 2001

 

题意：求子矩阵的最大和。

 

极度朴素的算法是O(N⁶)，加上预处理出和能达到O(N⁴)，但是这都不能达到要求。

我们在O(N⁴)的算法上加优化，可以少掉一层循环。

我们先来看一下这道题的线性版本。

给出一个序列，求其中一段连续子序列的最大和。

很明显我们可以预处理出前 i 个元素的和。然后 j ~ i 这段元素的和就是 sum[i] - sum[j - 1]，现在我们要找这个值的最大值，我们看到在 i 不变的时候，

如果我们想让这个值最大，那sum[j - 1]必定是 1 ~ i - 1最小的。

这道题只是把它扩充成二维的版本。

加上一点东西，把这个矩阵压缩成一条线就行了。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int MAXN = 105;
const int INF = 0x7fffffff;
int best;
int rowsum[MAXN][MAXN];
int sum[MAXN];
int n;
int ans;

int main()
{
    scanf("%d", &n);
    memset(rowsum, 0, sizeof(rowsum));
    for (int i = 1; i <= n; ++i)
    {
        for (int j = 1; j <= n; ++j)
        {
            scanf("%d", &rowsum[i][j]);
            rowsum[i][j] += rowsum[i][j - 1];
        }
    }
    ans = -INF;
    sum[0] = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = i; j <= n; ++j)
        {
            best = 0;
            for (int k = 1; k <= n; ++k)
            {
                sum[k] = rowsum[k][j] - rowsum[k][i - 1] + sum[k - 1]; /* 压缩 */
                ans = max(ans, sum[k] - sum[best]);
                if (sum[best] > sum[k]) best = k;
            }
        }
    printf("%d
", ans);
    return 0;
}