poj 1050 To the Max最大子矩阵和

最新推荐文章于 2025-09-06 09:26:50 发布

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最新推荐文章于 2025-09-06 09:26:50 发布

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文章标签： python

原文链接：https://my.oschina.net/locusxt/blog/133654

本文介绍了一个算法问题：在一个给定的N*N二维数组中找到具有最大和的子矩阵，并给出了具体的实现代码。该问题通过枚举子矩阵边界并计算其和来解决。

2019独角兽企业重金招聘Python工程师标准>>>

/**
To the Max
Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 35771		Accepted: 18784

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.
E:最大子矩阵查看提交统计提问总时间限制: 1000ms内存限制: 65536kB
描述
已知矩阵的大小定义为矩阵中所有元素的和。给定一个矩阵，你的任务是找到最大的非空(大小至少是1 * 1)子矩阵。

比如，如下4 * 4的矩阵

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2

的最大子矩阵是

9 2
-4 1
-1 8

这个子矩阵的大小是15。
输入
输入是一个N * N的矩阵。输入的第一行给出N (0 < N <= 100)。再后面的若干行中，依次（首先从左到右给出第一行的N个整数，
再从左到右给出第二行的N个整数……）给出矩阵中的N2个整数，整数之间由空白字符分隔（空格或者空行）。已知矩阵中整数的范
围都在[-127, 127]。
输出
输出最大子矩阵的大小。
样例输入
4
0 -2 -7 0 9 2 -6 2
-4 1 -4  1 -1

8  0 -2样例输出
15
*/

#include <cstdio>
using namespace std;
int n = 0;
int matrix[105][105] = {0};

int msum(int a, int b)
{
    int deal[105] = {0}, ms[105] = {0};
    for(int i = 0; i < n; ++i)
    for(int j = a; j <= b; ++j)
    deal[i] += matrix[j][i];

    ms[0] = deal[0];
    for(int i = 1; i < n; ++i)
    ms[i] = (ms[i - 1] + deal[i] > deal[i])?ms[i - 1] + deal[i]:deal[i];
    int ma = ms[0];
    for(int i = 0; i < n; ++i)
    if(ma < ms[i]) ma = ms[i];
    return ma;

}


int main()
{
    scanf("%d", &n);
    for(int i = 0; i < n; ++i)
    for(int j = 0; j < n; ++j)
    scanf("%d", &matrix[i][j]);
    int m = -100000000;
    for(int i = 0; i < n; ++i)
    for(int j = i; j < n; ++j)
    {
        int tep = msum(i, j);
        if(m < tep) m = tep;
    }
    printf("%d\n", m);
}

转载于:https://my.oschina.net/locusxt/blog/133654