To the Max dp-动态规划
Time limit:1000 ms Memory limit:10000 kB Source: Greater New York 2001描述
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.
输入
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].
输出
Output the sum of the maximal sub-rectangle.
Input
4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2
Output
15
代码
#include<iostream>
#include<algorithm>
using namespace std;
const int MAX = 101;
int arr[MAX][MAX] = { 0 };
int Dy[MAX][MAX] = { 0 };
int main() {
std::ios::sync_with_stdio(false);
//freopen("input.txt", "r", stdin);
int N;
cin >> N;
for (int i = 1; i <= N; i++)
{
for (int j = 1; j <= N; j++)
{
cin >> arr[i][j];
Dy[i][j] = Dy[i][j - 1] + arr[i][j]; //让Dy存放第i行的前j项和
//其实我们用数学之美来看为什么要这样存放,你一维的时候存放的是一维的数据
//二维的时候再存放一维的数据就是错误的,二维的时候就应该存放二维的数据
//当然你也不要去存放前i行前j列之和,这个相比于前面的就是三维的数据了
//如果你算的是长方体的数据,这样应该是对的,那么我们的数组也应该是三维的了
}
}
int max = INT_MIN;
int sum;
for (int i = 1; i <= N; i++)
{
for (int j = i; j <= N; j++)
{
sum = 0;
//前面两层循环是列循环,比如三列的话就是,1-2列,1-3列,2-3列
for (int k = 0; k <= N; k++)
{
//行循环,求和
if (sum < 0) sum = 0;
sum += (Dy[k][j] - Dy[k][i - 1]);
if (sum > max) max = sum;
}
}
}
cout << max << endl;
}
本人也是新手,也是在学习中,勿喷
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