HDU 1081 To The Max

Problem Description

Given atwo-dimensional array of positive and negative integers, a sub-rectangle is anycontiguous 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. Inthis problem the sub-rectangle with the largest sum is referred to as themaximal 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 consistsof an N x N array of integers. The input begins with a single positive integerN 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 secondrow, left to right, etc. N may be as large as 100. The numbers in the arraywill be in the range [-127,127].

 

 

Output

Output the sum ofthe 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

 

求最大子矩阵的和。这道题里面没有这样的例子，全是负数的情况，输出0或者最大的一个负数都可以。

转移方程：dp[i][j][k] = max(sum(num[i][j]---num[i][k]) , dp[i-1][j][k] + sum(num[i][j]---num[i][k]))

dp[i][j][k]表示的是矩阵最后一行为第i行第j个元素到第k个元素最大值。这个矩阵只跟其上一层为最后一行的矩阵有关系。

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

int num[110][110] ,n ,m;
int dp[110][110][110];//dp[i][j][k]表示矩阵最后一行为第i行第j个元素到第k个元素最大值

int main()
{
	int sum;
	while(~scanf("%d",&n))
	{
		for(int i = 1;i <= n;i++)
		{
			for(int j = 1;j <= n;j++)
			{
				scanf("%d",&num[i][j]);
			}
		}
		m = INT_MIN;
		for(int i = 1;i <= n;i++)
		{
			for(int j = 1;j <= n;j++)
			{
				for(int k = j;k <= n;k++)
				{
					sum = 0;
					for(int s = j;s <= k;s++)
					{
						sum += num[i][s];
					}
					dp[i][j][k] = sum;
					if(dp[i][j][k] < dp[i - 1][j][k] + sum)
					{
						dp[i][j][k] = dp[i - 1][j][k] + sum;
					}
					if(m < dp[i][j][k])
					{
						m = dp[i][j][k];
					}
				}
			}
		}
		printf("%d\n",m);
	}
	return 0;
}

求第j到k个元素的和也可以继续用dp的思想。同时，因为从第一行依次往下，且每次只会用到当前行和前一行。因此，只用记录前一行j到k之和就可以了。

sum用来保存当前行j到k之和，sum_front记录前一行j到k之和。dp用来记录当前行为最后一行j到k的矩阵的最大值。

这个做法较上一个做法减少了算j到k之和的时间，没有暴力计算。

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

int num[110][110] ,n ,m;
int dp[110][110] ,sum[110][110];

int main()
{
	int sum_front;
	while(~scanf("%d",&n))
	{
		memset(dp,0,sizeof(dp));
		for(int i = 1;i <= n;i++)
		{
			for(int j = 1;j <= n;j++)
			{
				scanf("%d",&num[i][j]);
			}
		}
		m = 0;
		for(int i = 1;i <= n;i++)
		{
			for(int j = 1;j <= n;j++)
			{
				for(int k = j;k <= n;k++)
				{
					sum_front = dp[j][k];
					sum[j][k] = sum[j][k-1] + num[i][k];
					dp[j][k] = sum[j][k];
					if(0 < sum_front)
					{
						dp[j][k] += sum_front;
					}
					if(m < dp[j][k])
					{
						m = dp[j][k];
					}
				}
			}
		}
		printf("%d\n",m);
	}
	return 0;
}