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）：

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #include <cstring>

 using namespace std;

 const int MAXN = ;

 int mat[MAXN][MAXN], n;

 int sum[MAXN][MAXN];

 void calsum() {

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

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

 }

 int solve() {

     int ans = -;

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

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

             int t = ;

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

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

                 ans = max(t, ans);

                 if(t < ) t = ;

             }

         }

     }

     return ans;

 }

 int main() {

     scanf("%d", &n);

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

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

     calsum();

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

 }

巴特西

URAL 1146 Maximum Sum（DP）

Input

Output

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