POJ 1050 To the Max -- 动态规划

题目地址：http://poj.org/problem?id=1050

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

时间复杂度为O(N^2*M^2)     

#include <stdio.h>
#include <limits.h>

//PS[i][j]等于以(1, 1), (1, j), (i, 1), (i, j)为顶点的矩形区域的元素之和
void Preproccess(int matrix[101][101], int PS[101][101], int N){
	int i, j;
	for (i=0; i<=N; ++i){
		PS[0][i] = 0;
		PS[i][0] = 0;
	}
	for (i=1; i<=N; ++i){
		for (j=1; j<=N; ++j){
			PS[i][j] = PS[i-1][j] + PS[i][j-1] - PS[i-1][j-1] + matrix[i][j];
		}
	}
}

int main(void){
	int N;
	int matrix[101][101];
	int PS[101][101];
	int i, j;
	int i_min, i_max;
	int j_min, j_max;
	int max, tmp;

	while (scanf ("%d", &N) != EOF){
		for (i=1; i<=N; ++i)
			for (j=1; j<=N; ++j)
				scanf ("%d", &matrix[i][j]);
		Preproccess(matrix, PS, N);
		max = INT_MIN;
		/*以(i_min, j_min), (i_max, j_min), (i_min, j_max), (i_max, j_max)为顶点的矩形区域的元素之和，
		等于PS[i_max][j_max] - PS[i_min-1][j_max] - PS[i_max][j_min-1] + PS[i_min-1][j_min-1]
		*/
		for (i_min=1; i_min<=N; ++i_min){
			for (i_max=i_min; i_max<=N; ++i_max){
				for (j_min=1; j_min<=N; ++j_min){
					for (j_max=j_min; j_max<=N; ++j_max){
						tmp = PS[i_max][j_max] - PS[i_min-1][j_max] - PS[i_max][j_min-1] + PS[i_min-1][j_min-1];
						if (tmp > max)
							max = tmp;
					}
				}
			}
		}
		printf ("%d
", max);
	}

	return 0;
}

时间复杂度为O(N*M*min(N, M))

#include <stdio.h>
#include <limits.h>

//PS[i][j]等于以(1, 1), (1, j), (i, 1), (i, j)为顶点的矩形区域的元素之和
void Preproccess(int matrix[101][101], int PS[101][101], int N){
	int i, j;
	for (i=0; i<=N; ++i){
		PS[0][i] = 0;
		PS[i][0] = 0;
	}
	for (i=1; i<=N; ++i){
		for (j=1; j<=N; ++j){
			PS[i][j] = PS[i-1][j] + PS[i][j-1] - PS[i-1][j-1] + matrix[i][j];
		}
	}
}

//BC(PS, a, c, i)表示在第a行和第c行之间的第i列的所有元素的和，可以通过“部分和”PS[i][j]在O(1)时间内计算出来。
int BC(int PS[101][101], int a, int c, int i){
	return PS[c][i] - PS[a-1][i] - PS[c][i-1] + PS[a-1][i-1];
}

int MaxSum (int matrix[101][101], int PS[101][101], int N){
	int max = INT_MIN;
	int a, c, i;
	int Start, All;
	for (a=1; a<=N; ++a){
		for (c=a; c<=N; ++c){
			Start = BC(PS, a, c, N);
			All = BC(PS, a, c, N);
			for (i=N-1; i>=1; --i){
				if (Start < 0)
					Start = 0;
				Start += BC(PS, a, c, i);
				if (Start > All)
					All = Start;
			}
			if (All > max)
				max = All;
		}
	}
	return max;
}

int main(void){
	int N;
	int matrix[101][101];
	int PS[101][101];
	int i, j;
	
	while (scanf ("%d", &N) != EOF){
		for (i=1; i<=N; ++i)
			for (j=1; j<=N; ++j)
				scanf ("%d", &matrix[i][j]);
		Preproccess(matrix, PS, N);
		printf ("%d
", MaxSum (matrix, PS, N));
	}

	return 0;
}

HDOJ上相似的题目：http://acm.hdu.edu.cn/showproblem.php?pid=1559

九度OJ上相似的题目：http://ac.jobdu.com/problem.php?pid=1492

参考资料：编程之美