UVa 108：Maximum Sum

UVa 108：Maximum Sum2014-12-02链接：

http://uva.onlinejudge.org/index.php？option=com_onlinejudge&Itemid=8&category=113&page=show_problem&problem=44

原题：

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:

is in the lower-left-hand corner:

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

40 -2 -70 92 -62-41 -41 -180 -2

Sample Output

题目大意：

给一个N*N大小的矩阵，每个格子上有一个数字。求这个矩阵中的子矩阵上面数字之和最小的数字是多少。

分析与总结：

经典的一道题。也可以说是“最大连续和"那题（UVa 507 - Jill Rides Again） O（n）算法的升级版本。

这题的关键在于转化，就是把矩阵上面的那些数转换成”最大连续和“。

看下图（画得实在有点对不起观众 = =）

把那些格子看成数字，上面有红色斜线阴影的代表其中一个高度为2的子矩阵，我们可以把这两层看做是一层，上下层数字之和代表一个新的数字，就可以用“最大连续和”的O（n）方法求出斜线阴影部分最大的一个2*x的矩阵之和。

要求整个大矩阵的最大子矩阵数字和，就需要枚举所层数的情况，需要用两层for循环。加上里面那个O（n）的线性时间扫描，最终复杂度是O（n^3）.

代码：

/**UVa:108 - Maximum Sum *Time: 0.012s*Author: D_Double**/#include<iostream>#include<cstdio>#include<cstring>#define MAXN 102using namespace std;int arr[MAXN][MAXN],sum[MAXN][MAXN];int mx,sm;int main（）{int T,m,n,x,y;int i,j;while（~scanf（"%d",&n））{ memset（sum,0,sizeof（sum））; memset（arr,0,sizeof（arr））;for（i=1; i<=n; ++i）{for（j=1; j<=n; ++j） {scanf（"%d",&arr[i][j]）;// sum[i][j] 表示对角顶点为（1,1），（i,j）这个矩阵里面所有原数之和sum[i][j] = sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+arr[i][j]; }}int maxSum=-2147483645;for（int i=1; i<=n; ++i）{for（int j=0; j<i; ++j）{int t, min;min=0;for（int k=1; k<=n; ++k）{ // 线性时间扫描t=sum[i][k]-sum[j][k]-min;if（t>maxSum） maxSum=t;if（sum[i][k]-sum[j][k] < min）min=sum[i][k]-sum[j][k];}}}printf（"%d
", maxSum）;}return 0;}

附：这题的又一次升级版 UVa 10827 - Maximum sum on a torus

作者：csdn博客 shuangde800