POJ 1050（最大子矩阵和）

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 47107 Accepted: 24952

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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

15

分析：

一维数组最大子段和的推广，开始的两层for循环枚举子矩阵的 起始行 与 终止行 ，并将从起始行到终止行的元素按列加起来存到a数组里，这样就变成了一维数组求最大和了，即 tmp 数组的最大子段和，也就是求起始列和终止列，调用函数F即可求出来。有了起始行、终止行、起始列、终止列，子矩阵也就确定下来了，总的时间复杂度为O（n3）。

代码：

#include <cstdio>#include <cstdlib>#include <algorithm>#include <cstring>#include <cmath>#include <iostream>using namespace std;typedef long long int ll; const int maxn = 105;int N;int matrix[maxn][maxn];int tmp[maxn];ll F(int* a){ll sum = -9999999999;ll this_sum = 0;for(int i=0; i<N; i++){this_sum += a[i];if(this_sum>sum)sum = this_sum;if(this_sum<0)this_sum = 0;}return sum;}int main(){ll sum,Max=0;scanf("%d",&N);for(int i=0; i<N; i++)for(int j=0; j<N; j++)scanf("%d",&matrix[i][j]);for(int i=0; i<N; i++){     //行 memset(tmp,0,sizeof(tmp));for(int j=i; j<N; j++){    //行 for(int k=0; k<N; k++)tmp[k] += matrix[j][k];Max = max(Max,F(tmp));}}printf("%I64d\n",Max);return 0;}