hdu 1081 To The Max 线性DP 蛮不错的题

To The Max

Time Limit : 2000/1000ms (Java/Other)Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 30Accepted Submission(s) : 16

Problem Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous 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. 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 x 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

/*
题目大意：给定一个含有正数和负数的矩阵，求其子矩阵的和的最大值

一维的情况很简单，如何把一维的情况转化为二维情况呢？
例如，对于本题的测试数据：
我们可以每次任选几行，压缩成一行，这样就转化为了一维情况。
例如，我们求1~2行中的最大子矩阵：即矩阵高为2（1~2行），宽为1：4的矩阵，可以先把1~2行相加，得到9 0 -13 2，再求这个单行的最大子段，由此就可以求得1~2行的最大子矩阵。
*/

#include<stdio.h>
#include<string.h>
int max,n;
int map[105][105],b[105];
int dp()
{
	int i,sum=0,m=-1000;
	for(i=0;i<n;i++)
	{
		sum=sum+b[i];
		if(sum<0) sum=0;
		if(sum>m) m=sum;
	}
	return m;
	
}
int main()
{
	int i,j,k,t,ans;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=0;i<n;i++)
			for(j=0;j<n;j++)
				scanf("%d",&map[i][j]);
		    max=0;//一开始没有初始化 没法和后面的比较大小啊  所以WA了  所以要注意也谢细节问题
			for(i=0;i<n;i++)//从第几行开始
			{
				for(k=0;k<n;k++)//行数
				{
					if(i+k<n)
					{
						memset(b,0,sizeof(b));
						for(t=i;t<=i+k;t++)
						{
							for(j=0;j<n;j++)
							   b[j]+=map[t][j];
						}
						ans=dp();
						max=max>ans?max:ans;
					}
				}
			}
			printf("%d\n",max);	
	}
	return 0;
}