UVALive2288 POJ1050 HDU1081 ZOJ1074 To The Max【最大子段和+DP】

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48948		Accepted: 25895

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

Source

Greater New York 2001

Regionals 2001 >> North America - Greater NY

问题链接：UVALive2288 POJ1050 HDU1081 ZOJ1074 To The Max。

问题简述：参见上文。

问题分析：

这是一个计算最大子矩阵和的问题。

可以将该问题转化为计算最大子段和问题，是一个经典的动态规划问题。

在行上，采用穷尽搜索的方法来解决。

在列上，通过计算最大子段和来达到计算最大子矩阵的目的。

程序说明：数组b[]用于计算各行之和，其起始行从0到n-1（穷举法），实际的计算过程是动态的。

参考链接：HDU1003 Max Sum【最大子段和+DP】

题记：（略）

AC的C++语言程序如下：

/* UVALive2288 POJ1050 HDU1081 ZOJ1074 To The Max */

#include <iostream>
#include <limits.h>
#include <string.h>

using namespace std;

const int N = 100;
int a[N][N], b[N];

int main()
{
    int n, maxval;

    while(cin >> n) {
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
                cin >> a[i][j];

        maxval = INT_MIN;
        for(int i=0; i<n; i++) {
            memset(b, 0, sizeof(b));

            for(int j=i; j<n; j++) {
                int sum = 0;
                for(int k=0; k<n; k++) {
                    b[k] += a[j][k];

                    if(sum + b[k] > 0)
                        sum += b[k];
                    else
                        sum = b[k];

                    maxval = max(maxval, sum);
                }
            }
        }

        cout << maxval << endl;
    }

    return 0;
}