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.
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\n", 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\n", 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