POJ-1050-To the Max（DP）

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 51817		Accepted: 27379

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

一、原题地址

传送门

二、大致题意

    求给定大小的矩阵中的最大矩阵和。

三、思路

    求出矩阵每行的前缀和，记作dp[i][j]，那么dp[i][x]-dp[i][y]就是矩阵在第i行上第y列到第x列的和。这样就把问题转化为了求最大的子序列和。我们只要枚举dp[i][x]-dp[i][y]中的x和y就可以了。

四、代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<stack>
using namespace std;
const int inf = 0x3f3f3f3f;
#define LL long long int 
long long  gcd(long long  a, long long  b) { return a == 0 ? b : gcd(b % a, a); }




int a[105][105], dp[105][105], sum[105];
int n;
int main()
{
	memset(dp, 0, sizeof(dp));
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= n; j++)
		{
			scanf("%d", &a[i][j]);
			dp[i][j] = dp[i][j - 1] + a[i][j];
		}
	}
	int ans = -inf;
	for (int l = 1; l <= n; l++)
	{
		for (int r = l ; r <= n; r++)
		{
			sum[0] = 0;
			for (int k = 1; k <= n; k++)
			{
				sum[k] = dp[k][r] - dp[k][l-1];
			}
			for (int k = 1; k <= n; k++)
			{
				if (sum[k - 1] >= 0)sum[k] += sum[k - 1];
				else sum[k] = sum[k];
				ans = max(ans, sum[k]);
			}
		}
	}
	printf("%d\n", ans);
	getchar();
	getchar();
}