Ural-1146Maximum Sum-最大子矩阵

Time limit: 0.5 second	Memory limit: 64 MB

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. 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. A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array.
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-hand corner and has the sum of 15.
Input
The input consists of an N × 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 white-space (newlines and spaces). These N 2 integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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
The output is the sum of the maximal sub-rectangle.
Sample

input	output
4	15
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2

最大子矩阵和：以前做过一维的最大连续和，现在换成二维，而思路还是类似，不过要提前将二维的转化为一维，处理起来就是很方便，求出每一行的前缀和，然后枚举列，即可求出最大子矩阵和。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

int n;

int a[110][110];

int ans ; 

int main()
{
    while(~scanf("%d",&n))
    {
        memset(a,0,sizeof(a));

        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&a[i][j]);//前缀和
                a[i][j]+=a[i][j-1];
            }
        }

        ans = -10000000;

        for(int i=1;i<=n;i++)
        {
            for(int j=i;j<=n;j++)//枚举列每一次求第i列到第j列之间的最大和
            {
                int Max = -10000000;

                int sum = 0;

                for(int k=1;k<=n;k++)
                {
                    sum+=(a[k][j]-a[k][i-1]);

                    Max = max(Max,sum);

                    if(sum<0)
                    {
                        sum = 0;
                    }
                }

                ans = max(ans,Max);
            }
        }

        printf("%d
",ans);
    }
    return 0;
}