HDU 1081 To The Max(最大连续子序列和（二维）)

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3805    Accepted Submission(s): 1816

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

Source

Greater New York 2001

解题报告：题意：求一个二维矩阵的最大子矩阵的和（最大子段和的推广），方法就是把二维的转化为一维的，

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
const int  N= 110;
int a[N][N],sum[N];
int main()
{
    //freopen("1002.txt","r",stdin);
    int n,i,j,k,sum1,max,ans;
    while(scanf("%d",&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        ans=-99999;
        for(i=1;i<=n;i++)
        {
            memset(sum,0,sizeof(sum));
            for(j=i;j<=n;j++)//将二维压缩成一维的
            {
                for(k=1;k<=n;k++)
                {
                    sum[k]+=a[j][k];
                }
                max=-99999;
                sum1=0;
                for(k=1;k<=n;k++)//求最大值
                {
                    if(sum1>0)
                    {
                        sum1+=sum[k];
                    }
                    else
                    {
                        sum1=sum[k];
                    }
                    if(max<sum1)
                    {
                        max=sum1;
                    }
                }
                if(ans<max)
                {
                    ans=max;
                }
            }
        }
        printf("%d\n",ans);
    }
}