hdu1081To The Max

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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 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.
 
InputThe 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].
 
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


最大連續子串和：

有串s,用一個臨時變數t維護一個ans，若t>=0則加上s[i]可能會更好，於是有t+=s[i]，
若t<0,則變成s[i]可能會更好，於是有t=s[i]
每次若大於ans，更新即可

最大子矩陣和：

有矩陣s，用一個臨時數組a維護一個ans，每次a都記錄s連續i行矩陣的和，
這樣就可以當作最大連續子串處理，每次算出最大值更新ans，即可

#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