POJ 1050_To the Max

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

题意：求矩阵中子矩阵的最大矩阵和

解题思路：我们求一维序列中最大子序列用DP从前往后扫一遍即可，那么对于二维的数组我们要想方法转成一维数据来处理，这个是核心的思想，剩下的就是怎么把二维转为一维。我们知道我们只能在一个维度上DP，所以另一个维度就要穷举。

我们选择在X方向上DP而在Y方向上穷举：我们求出每一种Y方向上的DP值，然后取最大值。比如3x3的数据，我们要求出1行DP，1~2行DP，1~3行DP，2行DP，2~3行DP，3行DP，取最大值即可。

下面是2种代码基本思路一致，只是细节上第二种更省时

#include <iostream>using namespace std;#define N 102int num;int data[N][N];//源数据int temp[N];//压缩后临时一维数组int dp(){int max = -(1<<30);int out = temp[0];for (int i=1; i<num; i++ ){if (out < 0) out = temp[i];elseout += temp[i];if (max < out) max = out;}return max;}int main(){cin>>num;int out = -(1<<30);//输入矩形for (int i=0; i<num; i++)for (int j=0; j<num; j++)cin>>data[i][j];for (int i=0; i<num; i++){memset(temp, 0, sizeof(temp));//清空临时一维数组for (int j=i; j<num; j++){//求一维数组for (int k=0; k<num; k++)temp[k] += data[j][k];//求最大值int a = dp();if (out < a) out = a;}}cout<<out<<endl;return 0;}

将DP过程精简在main中，避免了不必要的计算

#include <iostream>using namespace std;#define N 102int num;int data[N][N];//源数据int temp[N][N];//int main(){cin>>num;int max = -(1<<30);int out = -(1<<30);//输入矩形for (int i=1; i<=num; i++)for (int j=1; j<=num; j++)cin>>data[i][j];for (int i=1; i<=num; i++)for (int j=1; j<=num; j++)temp[i][j] = temp[i][j-1] + data[i][j];//求出temp表for (int i=1; i<=num; i++)for (int j=i; j<=num; j++){int a = temp[1][j] - temp[1][i-1];out = a;for (int k=2; k<=num; k++){if (out > 0){out += temp[k][j] - temp[k][i-1];}else{out = temp[k][j] - temp[k][i-1];}if (out > a) a  = out;}if (a > max) max = a;}cout<<max<<endl;//system("pause");return 0;}