hdu-1081 To The Max (最大子矩阵和)
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http://acm.hdu.edu.cn/showproblem.php?pid=1081
To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11309 Accepted Submission(s): 5454
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
动态规划问题
可以将二维数组压缩成一维数组,求压缩后的数字最大子段和,即二维数组的最大子矩阵和。
将二维数组压缩:将第1行到第n行的每一列都加起来,即a[i][j]表示第j列前1-i行数的和,再求第i行最大子段和就可以了
#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int main(){ int n, a[110][110], i, j, k; while(cin>>n){ memset(a, 0, sizeof(a)); for(i=1; i<=n; i++){ for(j=1; j<=n; j++){ cin>>a[i][j]; a[i][j] += a[i-1][j];//a[i][j]表示第j列前i行数的和 } } int mmax = -1000000, sum; for(i=1; i<=n; i++){ for(k=1; k<=i; k++){ sum = 0; for(j=1; j<=n; j++){ sum += (a[i][j]-a[k-1][j]); if(sum > mmax) mmax = sum; if(sum < 0) sum = 0; } } } cout<<mmax<<endl; } return 0;}
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