【POJ-1050】To The Max（动态规划）

To the Max

Time Limit: 1000MS
Memory Limit: 10000K

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

题目大意

给定一个N*N的二位数组Matrix，求该二维数组的最大子矩阵和。

题目分析

暴力枚举

用4个循环，枚举出所有的子矩阵，再给每个子矩阵求和，找出最大的，肯定会超时，不可用。
时间复杂度：O(N^6)

动态规划（标准解法）

把二维转化为一维再求解。
有子矩阵：矩阵中第i行至第j行的矩阵。
用数组ColumnSum[k]记录子矩阵中第k列的和。
最后对ColumnSum算出最大子段和进行求解。
时间复杂度：O(N^3)

关于最大子段和：
有一序列a=a1 a2 ... an，求出该序列中最大的连续子序列。
比如序列1 -2 3 4 -5的最大子序列为3 4，和为3+4=7。
动态转移方程：DP[i]=max(DP[i-1]+a[i], a[i])
时间复杂度：O(N)
（最大子段和的具体过程网上有，我就不多说了）

代码

#include <cstdlib>
#include <cstdio>
using namespace std;
#define INF 0x7f7f7f7f
#define max(a, b) (((a)>(b))?(a):(b)) 
int N;
int Matrix[110][110];
int Answer = -INF; 
int main()
{
    scanf("%d", &N);
    for(int i = 1; i <= N; ++ i)
        for(int j = 1; j <= N; ++ j)
            scanf("%d", &Matrix[i][j]);
    for(int i = 1; i <= N; ++ i)
    {
        int ColumnSum[110] = {0};
        for(int j = i; j <= N; ++ j)
        {
            int DP[110] = {0}; 
            for(int k = 1; k <= N; ++ k)
            {
                ColumnSum[k] += Matrix[j][k]; 
                // 求最大子段和 
                DP[k] = max(DP[k-1] + ColumnSum[k], ColumnSum[k]);
                Answer = max(DP[k], Answer); 
            }
        }
    }
    printf("%d\n", Answer);
    return 0;
}