hdu1081 To The Max（最大子矩阵和）

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
4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Sample Output
15

大致题意：让你找出和最大的子矩阵。

思路：枚举最大子矩阵的上下界，然后将每一列的值相加存到一个数组中，然后问题就转化成了求最大连续子段和时间复杂度为O（n），总的时间复杂度为n^3

代码如下

#include<cstdio>  #include<cstring>  #include<algorithm> #include<iostream> using namespace std;  int a[105][105];int b[105];int main(){    std::ios::sync_with_stdio(false);    int n;    while(cin>>n)    {    for(int i=1;i<=n;i++)    for(int j=1;j<=n;j++)        cin>>a[i][j];    int ans=-1e8;    for(int i=1;i<=n;i++)    {        memset(b,0,sizeof(b));        for(int j=i;j<=n;j++)        {            for(int k=1;k<=n;k++)            b[k]+=a[j][k];            int sum=0;            for(int k=1;k<=n;k++)            {                if(sum<0)                    sum=b[k];                else                     sum+=b[k];                ans=max(ans,sum);            }        }           }    cout<<ans<<endl;        }    return 0;}