POJ - 1050 To the Max （最大子段和子矩阵和最大降维）

最新推荐文章于 2019-09-25 16:51:01 发布

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最新推荐文章于 2019-09-25 16:51:01 发布

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分类专栏： ACM水题文章标签：降维最大子段和

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题目链接：

http://poj.org/problem?id=1050

题目：

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

题目大意：

给你一个二维矩阵，让你求一个子矩阵，使得子矩阵的和最大。

题目分析：

首先，先想一下无长度限制的最大字段和问题如何解决0.0 只需O（n）扫描该数列，不断把新的数加入子段，当子段和变成负数时，就把当前子段清空。扫描过程中出现的最大子段和就是所求。

那么，二维的应该怎么做的。我们可以想办法把二维的压缩成一维的。怎么压缩呢？枚举i，j，使得temp【k】=a【i】+a【i+1】+.....+a【j】，这样，枚举所有的 i j 分别求解，记录最大值即可。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;
const int MAXN = 100 + 100;
int a[MAXN][MAXN], temp[MAXN];

int GetLineMaxSum(int *num, int n) {
    int ans = int(-1e9), temp = 0;
    for (int i = 1; i <= n; i++) {
        if (temp + num[i] >= 0) {
            temp += num[i];
            ans = max(temp, ans);
        } else
            temp = 0;
    }
    return ans;
}

int main() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            scanf("%d", &a[i][j]);
    int ans = (int) -1e9;
    for (int i = 1; i <= n; i++) {
        for (int j = i; j <= n; j++) {
            for (int k = 1; k <= n; k++) {
                temp[k] = 0;
                for (int l = i; l <= j; l++)
                    temp[k] += a[k][l];
            }
            ans = max(GetLineMaxSum(temp, n), ans);
        }
    }
    printf("%d\n", ans);
    return 0;
}