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

15

題目大意

給定一個N*N的二位數組Matrix，求該二維數組的最大子矩陣和。

題目分析

• 暴力枚舉

用4個循環，枚舉出所有的子矩陣，再給每個子矩陣求和，找出最大的，肯定會超時，不可用。
時間復雜度：O(N^6)

• 動態規劃 （標準解法）

把二維轉化為一維再求解。
有子矩陣：矩陣中第i行至第j行的矩陣。
用數組ColumnSum[k]記錄子矩陣中第k列的和。
最後對ColumnSum算出最大子段和進行求解。
時間復雜度：O(N^3)

關於最大子段和：
有一序列a=a1 a2 ... an，求出該序列中最大的連續子序列。
比如序列1 -2 3 4 -5的最大子序列為3 4，和為3+4=7。
動態轉移方程：DP[i]=max(DP[i-1]+a[i], a[i])
時間復雜度：O(N)
（最大子段和的具體過程網上有，我就不多說了）

代碼

#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;
} 

【POJ-1050】To The Max（動態規劃）