Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 8992 Accepted Submission (s): 4359
Problem Descriptiongiven 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 the "the elements in" that rectangle. The problem the sub-rectangle with the largest sum are 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.
Inputthe input consists of an n x n array of integers. The input begins with a single positive integer N in a line by itself, indicating the size of the square two-dimensional a Rray. This was followed by N 2 integers separated by whitespace (spaces and newlines). These is the N 2 integers of the array, presented in Row-major order. That's, all numbers on the first row, left-to-right, then all numbers-second row, left-to-right, etc. N may be as large as 100. The numbers in the array would be in the range [-127,127].
Outputoutput 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
Sourcegreater New York 2001
Maximum consecutive substrings and:
There is a string of s, with a temporary variable T to maintain a ans, if t>=0 is added s[i] may be better, so there is t+=s[i],
If t<0, then become s[i] may be better, so there is t=s[i]
If it is greater than ans each time, update can
Maximum sub-matrices and:
There is a matrix s, with a temporary array A to maintain an ans, each time A is recorded s continuous I row matrix and,
This can be used as the maximum continuous substring processing, each time the maximum value updated ans, you can
#include <map> #include <string> #include <cstring> #include <cstdio> #include <cstdlib># include<cmath> #include <queue> #include <vector> #include <iostream> #include <algorithm > #include <bitset> #include <climits> #include <list> #include <iomanip> #include <stack > #include <set>using namespace std;int a[110],pic[110][110];int n;int sub () {int ans=int_min,t=0;for (int i=0;i <n;i++) {if (t>=0) T+=a[i];elset=a[i];ans=max (ans,t);} return ans;} int matrix () {int ans=int_min;for (int i=0;i<n;i++) {memset (a,0,sizeof (a)), for (int j=i;j<n;j++) {for (Int. k=0;k <n;k++) A[k]+=pic[j][k];ans=max (Ans,sub ());}} return ans;} int main () {while (cin>>n) {for (Int. i=0;i<n;i++) for (int j=0;j<n;j++) cin>>pic[i][j];cout<< Matrix () <<endl;}}
Hdu1081to the Max
