POJ To the Max 1050 （矩阵求和）动态规划算法

问题描述

原题链接
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.

算法分析

动态规划算法，用a[i][j]表示第j列的1到i行元素的和，所以a[i][j]就组成了一个一维数组，转化成了求一维子段和的问题。设C[i]为以i结尾的最大子段和，那对于C[i]而言只有两种情况，如果C[i - 1] > 0, 那么C[i] = C[i - 1] + W[i];如果C[i - 1] <=0,C[i] = W[i]，然后求出C数组中的最大值即可。对应到这个问题上，W[i]= a[j][z] - a[i][z],所以如果b[z - 1] >= 0，b[z] = b[z - 1] + a[j][z] - a[i][z];如果b[z - 1] < 0,b[z] = a[j][z] - a[i][z];

代码实现

#include<iostream>  
#include<cstring>  
using namespace std;
int main() 
{
	int N, a[100][100] = {0}, b[100] = { 0 };
	int num = a[1][1];
	cin >> N;
	for (int i = 1; i <= N; ++i)
	{
		for (int j = 1; j <= N; ++j)
		{
			cin >> a[i][j];
			a[i][j] += a[i - 1][j];
		}
	}
	
	for (int i = 0; i <= N - 1; ++i)
	{
		for (int j = i + 1; j <= N; ++j)
		{
			for (int k = 1; k <= N; ++k)
			{
				if (b[k - 1] >= 0)
					b[k] = b[k - 1] + a[j][k] - a[i][k];
				else
					b[k] = a[j][k] - a[i][k];
				if (num < b[k])
					num = b[k];
			}
		}
	}
	cout << num << endl;
	return 0;
}