HDU - 1081 To The Max

To The Max

Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

Submit Status

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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 

 

关键在压缩，将矩阵压缩为一维数组，然后转变为求最大字段和 ，一维数组的元素为各个列从i行到j行的数据之和，在枚举i，j。即可求得答案。

#include<iostream>#include<cstring>using namespace std;int a[110][110]; int sum[110][110];     //sum[i][j]表示第i行从1到j的元素之和。int main(){int n;while (scanf("%d", &n) != EOF){for (int i = 1; i <= n;i++)for (int j = 1; j <= n; j++)scanf("%d", &a[i][j]);for (int k = 1; k <= n; k++){sum[k][0] = 0;for (int j = 1; j <= n; j++)sum[k][j] = sum[k][j - 1] + a[k][j];      //求出sum数组}int c;int Max=-10e6;for (int i = 1; i <= n;i++)for (int j = i; j <= n; j++){                                              c = 0;for (int k = 1; k <= n; k++)              //对列进行动态规划，与求最大连续子段和相同，但段的元素变成了 第k列 中“从第i行到第j行的数据之和”{                                         //该元素可由sum[k][j]-sum[k][i-1]得到。c = c + sum[k][j] - sum[k][i-1];if (c < 0) c = 0;if (c > Max) Max = c;}}cout << Max << endl;}}