poj1050 dp动态规划

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

题目翻译过来，就是在一个大矩阵里面，找一个（矩阵和最大）的子矩阵。

如题意：  
0 -2 -7 0 
9 2 -6 2          9  2
-4 1 -4 1  　　　　-4 1
-1 8 0 -2         -1 8       9+2+（-4）+1+（-1）+8　　=    =  15  所以输出15

做这题之前，先来了解一下　　　　一维数组子串，找连续数组的最大和           例如    2  3   -7  9 2  -6  9       最大和为（9,2，-6,9）   ->     14

如何用算法实现，当然用dp     顺推，   假设数组a    只要a的前项与后项的和大于0，保留，继续比较，在这个过程中要存储max的值     

正如     上面例子

  2  3   -7  9 2  -6  9    

遍历一遍   (前项加本项等于本项的值，但形成负数的时候，要归零)

　形成   2    5（2+3）    0 （5+-7小于0，归零）  9  　　 11（9+2）  　　5 （11-6）　　 14（5+9）

同样本题是这个思想的延伸，扩展到了二维

第一步：
0 -2 -7 0 　　　　->  max=0
9 2 -6 2 　　　　 ->　max=11    
-4 1 -4 1 　　　　->  max=1         四个合起来 ，得max=11
-1 8 0 -2 　　　　->  max=8

 第二步（二维压缩成一维 ）：

第一行与第二行相加保留在第二行，然后继续压缩，最后压缩成了一维。中间过程要与max比较，取较大值.........


AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#define Max(a,b) ((a)>(b)?:(a):(b))
#define Min(a,b) ((a)<(b)?:(a):(b))
#define Mem0(x) memset(x,0,sizeof(x))
#define Mem1(x) memset(x,-1,sizeof(x))
#define MemX(x) memset(x,0x3f,sizeof(x))
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
int n,a[1010][1010];
/*
int dp(int b[])
{
	int i,ans=b[0];
	for (i=0;i<n;i++)
		printf("%d\t",b[i]);
	printf("******");
	for (i=0;i<n;i++){
		if (b[i]>0)
			b[i+1]+=b[i];
		if (ans<b[i])
			ans=b[i];
		printf("%d\t",b[i]);
	}
	printf("%d\t",b[i]);
	return ans;
}*/
int main()
{
	int n;
	while (cin>>n){
	int i,j,k,ans=-inf;
	Mem0(a);
	for (i=0;i<n;i++){
		int tmp=0;
		for (j=0;j<n;j++){
			cin>>a[i][j];
			if (tmp>0)
				tmp+=a[i][j];
			else tmp=a[i][j];
			if (tmp>ans)
				ans=tmp;
		}
	}
	for (i=0;i<n-1;i++){
		for (j=i+1;j<n;j++){
			int tmp=0;
			for (k=0;k<n;k++){
				a[i][k]+=a[j][k];
				if (tmp>0)
					tmp+=a[i][k];
				else
					tmp=a[i][k];
				if (tmp>ans)
					ans=tmp;
			}
		}
	}
	printf("%d\n",ans);
	}

//	system("pause");
	return 0;
}