杭电1081-To The Max

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6512    Accepted Submission(s): 3117

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.

 

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
这题可以（On^4）暴力过，但是这是数据较小，如果数据大就会TE了，所以我是用矩阵压缩和求最大连续子序列方法
/*子矩阵必定也是由行和列组成,如上这个矩阵它的行的组合有1,1-2,1-3,1-4，2,2-3,2-4，3，3-4,4无非这10种组合这样我们就可以将行进行压缩比如说：1-2我们将1,2行数据进行压缩及进行合并   0 -2 -7 0 + 9  2 -6 2-------------  9 0  -13 2那么这样我们就可以通过求最大连续子串和的思想求其最大值为9再看：2-4这个组合     9 2 -6  2    -4 1 -4  1 +  -1 8  0 -2----------------    4  11 -10 1所以它的最大和为15这样通过压缩行即将这个矩形的宽变为1可以迅速的求出子矩阵的值并求出最大值因为求最大连续子串和思想是线性的复杂度为O(N)所以可以快速求出压缩后矩阵的和！*/#include<iostream>#include<cstring>#include<string>#include<cstdio>#include<cstdlib>#include<cmath>#include<algorithm>#include<queue>#include<stack>#include<iomanip>#include<map>#include<set>const int MAX=101;int matrix[MAX][MAX];int DP[MAX][MAX];using namespace std;int main(){    int n,i,j,Max,maxn,sum;    while(cin>>n)    {        for(i=1;i<=n;i++)        {            for(j=1;j<=n;j++)            {                cin>>matrix[i][j];            }        }        memset(DP,0,sizeof(DP));        for(i=1;i<=n;i++)        {            for(j=1;j<=n;j++)            {                DP[j][i]=DP[j][i-1]+matrix[i][j];//dp[j][i]代表第j列从第1行开始的数累加到到第i行的和            }        }        Max=0;        for(i=1;i<=n;i++)        {            for(j=i;j<=n;j++)            {                maxn=0;                sum=0;                for(int k=1;k<=n;k++)                {                    sum+=DP[k][j]-DP[k][i-1];//表示第k列从第i行到第j行数字之和                    if(sum>0)                    {                        if(sum>maxn)                        maxn=sum;                    }                    else                    sum=0;                }                if(maxn>Max)                Max=maxn;            }        }        cout<<Max<<endl;    }    return 0;}