URAL 1146 Maximum Sum（DP）

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. 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. A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array.

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-hand corner and has the sum of 15.

Input

The input consists of an  N ×  N array of integers. The input begins with a single positive integer  Non a line by itself indicating the size of the square two dimensional array. This is followed by  N  2integers separated by white-space (newlines and spaces). These  N  2 integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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

The output is the sum of the maximal sub-rectangle.

 

题目大意：给一个n*n的矩阵，求和最大的子矩阵。

思路：用sum[i][j]表示从mat[1][j]~mat[i][j]的总和（从1开始计数）

然后枚举上下两行夹着的矩阵，设第一行为r1，第二行为r2，复杂度为O(n^2)，然后计算这两行夹着的最大子矩阵。

用sum[r2][j] - sum[r1 - 1][j]表示mat[r1][j]~mat[r2][j]的总和。

那么，我们把r1~r2之间的列，每一列算出来，就变成了一个只有n个元素的一维数组，求最大连续子序列。

这个就是经典问题了，设a[i] = sum[r2][i] - sum[r1][i]，初始化t = 0。

t从a[1]加到a[n]，当t ＜ 0的时候，令t = 0，算到 i 的时候，t就表示以a[i - 1]为结尾的最大后缀。

因为，如果我们算到a[i]，此时t ＜ 0，那么，算a[i + 1]的时候，肯定不会加上a[i]和前面的数字，不管怎么加，前面的数都小于0，还是不加的好。

能加的肯定要加上，所以复杂度为O(n)。

总复杂度为O(n^3)

 

代码（0.031S）：

 1 #include <cstdio>

 2 #include <iostream>

 3 #include <algorithm>

 4 #include <cstring>

 5 using namespace std;

 6 

 7 const int MAXN = 110;

 8 

 9 int mat[MAXN][MAXN], n;

10 int sum[MAXN][MAXN];

11 

12 void calsum() {

13     for(int i = 1; i <= n; ++i)

14         for(int j = 1; j <= n; ++j) sum[i][j] = sum[i - 1][j] + mat[i][j];

15 }

16 

17 int solve() {

18     int ans = -999;

19     for(int r1 = 1; r1 <= n; ++r1) {

20         for(int r2 = r1; r2 <= n; ++r2) {

21             int t = 0;

22             for(int j = 1; j <= n; ++j) {

23                 t += sum[r2][j] - sum[r1 - 1][j];

24                 ans = max(t, ans);

25                 if(t < 0) t = 0;

26             }

27         }

28     }

29     return ans;

30 }

31 

32 int main() {

33     scanf("%d", &n);

34     for(int i = 1; i <= n; ++i)

35         for(int j = 1; j <= n; ++j) scanf("%d", &mat[i][j]);

36     calsum();

37     printf("%d\n", solve());

38 }

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