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.
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
转化为一维数组的最大连续子串和。
#include<iostream>
#include<cstring>
using namespace std;
int line[101];
int matrix[101][101];
int n;
void input(void) {
cin >> n;
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
cin >> matrix[i][j];
}
int dp(void) {//一维数组求最大子串和
int tmp = 0;
int max = -(1 << 30);
for (int i = 0; i < n; i++) {
tmp += line[i];
if (tmp > max)
max = tmp;
if (tmp < 0)
tmp = 0;
}
return max;
}
int find(void) {
int max = -(1 << 30);
for (int i = 0; i < n; i++) {//i表示开始行
memset(line, 0, sizeof (line)); //每次开始数组清零
for (int j = i; j < n; j++) {//j表示结束行
for (int k = 0; k < n; k++)//k表示列
line[k] += matrix[j][k]; //将二维矩阵转化为一维数组
int t = dp();
if (t > max)
max = t;
}
}
return max;
}
int main(void) {
input(); //读入数据
cout << find() << endl; //输出答案
return 0;
}