POJ1050_To the Max_求最大子矩阵_DP

/*

吐槽：

唔。。。从早上就开始看这个题了- -又荒废了一天。

早上睡觉到十点半，然后就开了个会到12点，回来睡觉到4点多起来看B站，看楼教主的回忆录，跟看小说似的

到晚上才开始编

又是自己想了想觉得暴力不行就去搜代码了

- -不解释了

*/

题意：

求一个N*N的矩阵的最大子矩阵

思路：

把二维压缩成一维，转化为求一维数组的最大子序列（表示连一维的也木有做过，）但是此题只需求最大子矩阵的和值就行了，不需要知道位置，所以一维的dp甚至可以转化为0维！

下面是在别人博客里面找到的一维求最长子序列的方法：

（转载于http://blog.youkuaiyun.com/nomad2/article/details/6443620）

1、首先考虑一维的最大子段和问题，给出一个序列a[0]，a[1]，a[2]...a[n]，求出连续的一段，使其总和最大。

a[i]表示第i个元素
dp[i]表示以a[i]结尾的最大子段和

dp[i] = max{a[i], dp[i-1] + a[i]}

解释一下方程:

如果dp[i-1] > 0，则 dp[i] = dp[i-1] + a[i]
如果dp[i-1] < 0，则 dp[i] = a[i]

因为不用记录位置信息，所以dp[]可以用一个变量dp代替:

如果dp > 0，则dp += a[i]
如果dp < 0，则dp = a[i]

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二维转一维就可以暴力了，这里把第i行到第i+k行的数相加得到一维数组

原题：

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36381		Accepted: 19108

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

Run ID	User	Problem	Result	Memory	Time	Language	Code Length	Submit Time
11765976	chengtbf	1050	Accepted	208K	16MS	C++	1002B	2013-07-11 20:17:29

代码：

#include<stdio.h>
#include<string.h>
#define INF -10000000
#define N 105
int a[N][N],str[N],dp;//dp甚至可以降维到一个变量
int ans,n;

int find_max(int a,int b)
{
	if (a>b)
	{
		return a;
	}
	else
	{
		return b;
	}
}
void find_ans()//一维的求最长子序列
{
	int i;
	dp=0;//初始化dp，涵义是以i为结束的最长子串
	for ( i = 1; i <=n; i++)
	{
		dp=find_max(str[i],dp+str[i]);
		ans=find_max(dp,ans);
	}
}

int main()
{
	int i,j,i_start,i_end;
	while (scanf("%d",&n)!=EOF)
	{
		memset(a,0,sizeof(a));
		for ( i =1; i <=n ; i++)
		{
			for (  j= 1;  j<=n; j++)
			{
				scanf("%d",&a[i][j]);
				if(i>=2)
				{
					a[i][j]+=a[i-1][j];//a[i][j]表示当前第j列前i行的和a[1][j]+a[2][j]+....+a[i][j]
				}
			}
			
		}
		ans=a[1][1];//初始化结果
		for ( i_start = 1; i_start <=n ; i_start++)
		{
			for ( i_end = i_start; i_end <=n ; i_end++)
			{
				for ( j = 1; j <=n ; j++)
				{
					str[j]=a[i_end][j]-a[i_start][j];//初始化一维数组
				}
				find_ans();
			}
		}


		printf("%d\n",ans);

	}
	return 0;
}