hdu 1081 & poj 1050 To The Max(最大和的子矩阵)
2024-09-03 01:42:24
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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.
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].
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
题意:给你一个N*N的矩阵,求当中和最大的子矩阵的值!
代码例如以下:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int main()
{
int i, j, k, t;
int a, sum, max, N, m[147][147];
while(~scanf("%d",&N))
{
memset(m,0,sizeof(m));
for(i = 1; i <= N; i++)
{
for(j = 1; j <= N; j++)
{
scanf("%d",&a);
m[i][j]+=m[i][j-1]+a;//表示第i行前j个数之和
}
}
max = -128;
for(i = 1; i <= N; i++)//起始列
{
for(j = i; j <= N; j++)//终止列
{
sum = 0;
for(k = 1; k <= N; k++)//对每一行进行搜索
{
if(sum < 0)
sum = 0;
sum+=m[k][j]-m[k][i-1];
//m[k][j]-m[k][i-1]表示第k行第i列之间的数
if(sum > max)
max = sum;
}
}
}
printf("%d\n",max);
}
return 0;
}
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