To The Max

To The Max

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 10   Accepted Submission(s) : 7

Problem Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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.

 

Input

The input consists of an N x 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

Output the sum of the maximal sub-rectangle.

 

Sample Input

40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 0 -2

 

Sample Output

15

 

Source

Greater New York 2001

 

题目大意：

第一行输入一个数字N，然后构建N*N大小，且带有数值的矩阵。求出在这大矩阵中的小矩阵，要求数值和最大的。

差不多是第一题目的升级版吧，需要把为二维的思想扩展到二维上。所以，我设置了一个Now[]数值来实现维度的转变。

Now[]是用来存储每次从上到下，依次累加纵行的值，把二维转化成一维，然后再从Now[]中寻找连续和最大的值，既为相关区域中矩阵的和的最大值。依次寻找大矩阵中的小矩阵，求得最大数值和。

算法复杂度：

                        O（n^3）（大概吧...）

 

状态转移方程：

                        Now[i]=NUM[i][1]+NUM[i][2]+NUM[i][3]+.......+NUM[i][N];(i=1,2,3,4,.....N)

                        Max_S=Max（Sign，Sign+Now[i]）

                        Max=Max（Max，Max_S）

#include <stdio.h>
#include <string.h>
#include <string.h>
int main()
{
    int i,j,N,Max,sum,k,Sign,ii,Max_s;
    int NUM[105][105];          /*NUM[]是用来保存存入的大矩阵数值的数据*/
    int Now[105];                /*Now[]是用来实现维度的扩展,用来存储纵行的累加值*/   
    while(scanf("%d",&N)!=EOF)      /*输入矩阵大小*/
    {
        for(i=1;i<=N;i++)
            for(j=1;j<=N;j++)
                scanf("%d",&NUM[i][j]);     /*输入N*N大小的矩阵的数值*/
        for(i=1,Max=0;i<=N;i++)             /*i为Now[]的起始位置，从上到下，需要初始化Now[]N次操作*/
        {
            memset(Now,0,sizeof(Now));      /*每次从新的一行重新累加，需要重新初始化Now[]*/
            for(j=i;j<=N;j++)
            {
                for(k=1;k<=N;k++)
                    Now[k]+=NUM[j][k];      /*获取新的一行中的累加值*/
                for(ii=1,Max_s=0,Sign=0; ii<=N; ii++)
                {
                    if(Sign>0)              
                        Sign+=Now[ii];
                    else
                        Sign=Now[ii];
                    if(Max_s<Sign)      /*寻找Now中连续的最大值，既为某一区域的矩阵中的最大值，*/                
                        Max_s=Sign;     /*把每次找到的最大值存储在Mas_S中， */
                }
                if(Max<Max_s)           /*把本趟循环中的最大值与从开始到之前的最大值做比较，看是否替换最大值*/
                    Max=Max_s;
            }
        }
        printf("%d\n", Max);        /*全部循环结束输入，输出存储的最大值，即为大矩阵区域中，数值和最大的小矩阵*/
    }
    return 0;
}