(线性dp 最大子段和 最大子矩阵和)POJ1050 To the Max

时间:2022-01-17 17:01:22
To the Max
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 54338   Accepted: 28752

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

就是最大字段和的升级版,
从:http://www.cnblogs.com/fll/archive/2008/05/17/1201543.html 可知:

假设最大子矩阵的结果为从第r行到k行、从第i列到j列的子矩阵,如下所示(ari表示a[r][i],假设数组下标从1开始):
  | a11 …… a1i ……a1j ……a1n |
  | a21 …… a2i ……a2j ……a2n |
  |  .     .     .    .    .     .    .   |
  |  .     .     .    .    .     .    .   |
  | ar1 …… ari ……arj ……arn |
  |  .     .     .    .    .     .    .   |
  |  .     .     .    .    .     .    .   |
  | ak1 …… aki ……akj ……akn |
  |  .     .     .    .    .     .    .   |
  | an1 …… ani ……anj ……ann |

那么我们将从第r行到第k行的每一行中相同列的加起来,可以得到一个一维数组如下:
 (ar1+……+ak1, ar2+……+ak2, ……,arn+……+akn)
 由此我们可以看出最后所求的就是此一维数组的最大子断和问题,到此我们已经将问题转化为上面的已经解决了的问题了。

就是先让i从0到n遍历,然后j从i到n遍历,最后在第j行中k从0到n遍历,用一个数组分别保存每个列的各行的数字之和,就可以化为最大连续和(降维)。

复杂度为:O(n^3)

C++代码:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = ;
int d[maxn][maxn];
int s[maxn];
int INF = -0x3f3f3f3f; int MaxArray(int a[],int n){
int m = INF;
int tmp = -;
for(int i = ; i < n; i++){
if(tmp > )
tmp += a[i];
else
tmp = a[i];
if(tmp > m)
m = tmp;
}
return m;
} int main(){
int n;
scanf("%d",&n);
for(int i = ; i < n; i++){
for(int j = ; j < n; j++){
scanf("%d",&d[i][j]);
}
}
int ans = INF,tmp;
for(int i = ; i < n; i++){
memset(s,,sizeof(s));
for(int j = i; j < n; j++){
for(int k = ; k < n; k++){
s[k] += d[j][k];
}
tmp = MaxArray(s,n);
if(tmp > ans)
ans = tmp;
}
}
printf("%d\n",ans);
return ;
}