To The Max
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To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7484 Accepted Submission(s): 3625
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.
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/*首先 , 读数据的时候 a[i][j] 为前 i 行数据的第 j 个数的和 , 当求第 p 行到第 q 行之间宽为 p-q+1 的最大和矩阵的时候 , 就以 p 行和 q 行之间的同一列的数字的和作为一个数列的元素 , 然后 DP 求这个数列的最大和连续子序列 , 就是第 p 行到第 q 行之间宽为 p-q+1 的最大和矩阵 . */代码如下:#include <iostream>#include<string.h>using namespace std;int f[101][101];int main(){ int n,i,j,sum,x,y,max; while(cin>>n) { memset(f,0,sizeof(f)); for(j=1;j<=n;j++) for(i=1;i<=n;i++) { cin>>f[j][i]; f[j][i]+=f[j-1][i]; // if(f[j][i]>sum) sum=f[j][i]; } max=-127; for(i=1;i<=n;i++)for(j=i;j<=n;j++) { sum=0; for(y=1;y<=n;y++) { sum+=f[j][y]-f[i-1][y]; if(sum>max) max=sum; if(sum<0) sum=0; } } cout<<max<<endl; } return 0;}
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