To The Max(最大子矩阵问题)
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To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9268 Accepted Submission(s): 4495
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
这个是动态规划的最大子矩阵,跟最大子段的区别就是它是二维
的!所以要把它压缩成一维!这里的意思就是先算出每一列,再
算出每一行
这一题就是将一维的最大字段和扩展到二维,在一维的求最大字段和的过程中是这样操作的:
int max_sum(int n){ int i, j, sum = 0, max = -10000; for(i = 1; i <= n; i++) { if(sum < 0) sum = 0; sum += a[i]; if(sum > max) max = sum; } return sum;}
扩展到二维的时候也是同样的方法,不过需要将二维压缩成一维,所以我们要将数据做一下处理,使得map[i][j]从表示第i行第j个元素变成表示第i行前j个元素和,这样map[k][j]-map[k][i]就可以表示第k行从i->j列的元素和。只要比一维多两层循环枚举i和j就行了。
#include<cstdio>#include<iostream>#include<cstring>using namespace std;int map[105][105];int maxs;int main(){ int n,i,j,v,k,sum; while(cin>>n) { memset(map,0,sizeof(map)); for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { cin>>v; map[i][j]=map[i][j-1]+v; } }//这个的每一列的每一个数字是前面的数字之和!这样就确保了每一列是线性的! maxs=-100000; for(i=1;i<=n;i++) { for(j=i;j<=n;j++) { sum=0; for(k=1;k<=n;k++) { sum+=map[k][j]-map[k][i-1];//每一行再进行线性就可以实现二维数组的动态规划! if(sum<0) { sum=0; } if(sum>maxs) { maxs=sum; } } } } cout<<maxs<<endl; }}
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