poj1050TotheMax

最新推荐文章于 2021-02-28 23:11:31 发布

转载最新推荐文章于 2021-02-28 23:11:31 发布 · 299 阅读

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#numbers #whitespace #integer #input #output

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本文探讨如何在二维数组中找到具有最大和的子矩形，并提供了算法实现及示例解析。

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.

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 the sum of the maximal sub-rectangle.

0 -2 -7 0

9 2 -6 2

-4 1 -4 1

-1 8 0 -2

output

这个题目应该连着我上一次的那个“ 最大子字段和”的文章

地址是http://blog.youkuaiyun.com/dumpling5232/article/details/7757777

矩阵是二维的，压缩成一维的，然后求最大子字段和就可以了。

设sum[i][j] 表示第j列的，从第 1行累加到第i行的结果。

循环分三层

第一层 i:1-n表示从第i行开始加

第二层j:i-n表示累加到第j行

第三层 k:1-n 表示从第k列开始求最大子字段和

代码

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;
int ans[110];
int map[110][110];
int sum[110][110];
int dp[110];
int maxm;
int main()
{
    int n,i,j,k;
    while(scanf("%d",&n)!=EOF)
    {  for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
          scanf("%d",&map[i][j]);
          maxm=map[1][1];
       for(i=1;i<=n;i++) sum[1][i]=map[1][i];
       for(i=2;i<=n;i++)
       for(j=1;j<=n;j++)
         sum[i][j]+=sum[i-1][j]+map[i][j];

         for(i=1;i<=n;i++)
         {
             for(j=i;j<=n;j++)
             {
                 dp[1]=sum[j][1]-sum[i-1][1];
                 for(k=2;k<=n;k++)
                 {
                     int temp=sum[j][k]-sum[i-1][k];
                     dp[k]=max(dp[k-1]+temp,temp);
                     if(maxm<dp[k]) maxm=dp[k];
                 }
             }
         }
         printf("%d\n",maxm);
    }
  return 0;
}