Uva 507 最大子矩阵和

最新推荐文章于 2024-04-24 23:16:35 发布

nielaishan

最新推荐文章于 2024-04-24 23:16:35 发布

阅读量371

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM

本文链接：https://blog.youkuaiyun.com/NLSQQ/article/details/51991810

ACM 专栏收录该内容

62 篇文章

订阅专栏

本文介绍了一个经典的最大子矩阵和问题，即在一个二维数组中找到具有最大和的子矩形。文章提供了详细的解决思路和算法实现，利用一维最大子段和的方法进行求解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A - Maximum Sum

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Submit Status

Description

Background

A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem.

The Problem

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. 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. A sub-rectangle is any contiguous sub-array of size or greater located within the whole array. As an example, the maximal sub-rectangle of the array:

displaymath35

is in the lower-left-hand corner:

displaymath37

and has the sum of 15.

Input and Output

The input consists of an 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 integers separated by white-space (newlines and spaces). These integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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].

The output is 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

来源： <http://acm.hust.edu.cn/vjudge/contest/view.action?cid=99978>

题意：

最大子矩阵和的问题，期间注意sum的初始化和b的变化，当b>0的时候

b+=a[j];b<=0时 b=a[j];

方法：把i行到j行的上下行的数据相加。应用一维的最大字段和方法。

代码：

     
      
       #include <iostream>
      
      
       #include <cstdio>
      
      
       #include <cstring>
      
      
       #include <queue>
      
      
       #include <stack>
      
      
       #include <string>
      
      
       #include <set>
      
      
       #include <map>
      
      
       #include <cstdlib>
      
      
       #include <cmath>
      
      
       #include <algorithm>
      
      
       #define MAXX 100+10
      
      
       #define inf 0x3f3f3f3f
      
      
       using namespace std;
      
      
        
      
      
       int maxsum(int *a, int n)
      
      
       {
      
      
           int b=0, sum=-2147483645;
      
      
           for (int i=0; i<n; i++)
      
      
           {
      
      
               if (b<0)
      
      
                   b = a[i];
      
      
               else
      
      
                   b += a[i];
      
      
               if (b>sum)
      
      
                   sum = b;
      
      
           }
      
      
           return sum;
      
      
       }
      
      
        
      
      
       int main()
      
      
       {
      
      
           #ifdef LOCAL
      
      
               freopen("in.txt", "r", stdin);
      
      
               //freopen("out.txt", "w", stdout);
      
      
           #endif // LOCAL
      
      
           int n;
      
      
           while (scanf("%d", &n)!=EOF)
      
      
           {
      
      
               int a[MAXX][MAXX], b[MAXX];
      
      
               for (int i=0; i<n; i++)
      
      
                   for (int j=0; j<n; j++)
      
      
                   scanf("%d", &a[i][j]);
      
      
               int ans = -2147483645;
      
      
               for (int i=0; i<n; i++)
      
      
               {
      
      
                   memset(b, 0, sizeof(b));
      
      
                   for (int j=i; j<n; j++)
      
      
                   {
      
      
                       for (int k=0; k<n; k++)
      
      
                           b[k] += a[j][k];
      
      
                       int maxx = maxsum(b, n);
      
      
                       if (maxx > ans)
      
      
                           ans = maxx;
      
      
                   }
      
      
               }
      
      
               printf("%d\n", ans);
      
      
           }
      
      
           return 0;
      
      
       }
      
      
       

      
      
       相关的知识： 
       http://blog.youkuaiyun.com/liufeng_king/article/details/8632430