hdu 1081 To The Max ( 最大子矩阵 )

To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6612 Accepted Submission(s): 3159

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 

题意：

求矩阵里面的>=1*1规模的子矩阵的最大sum和。sum为子矩阵中各元素的和。

分析：

这题相当于是求二维的最大子串和，想想当初做hdu 1003 max sum一维的时候，就是看当前sum+第i个数,

如果sum<0,则sum=0并以第i+1个数为第一个数继续累加。每次为负就重复这个过程。

现在想想，其实就是一个状态转移的公式：

f[i]=max(0,f[i-1])+a[i]

至于这题，开始找最优子结构的是时候处理错了，开始想的是先找4个为第一层最小结构，然后再接着扩展。

但是这样很容易漏掉一些情况，且对数据处理也很麻烦，甚至到最后自己都搞晕了。

正确的思路是：

先把矩阵预处理。使得map[i][j]存的是第i行前j个数的和。

例如：样例中矩阵

           0 -2  -7 0

           9  2  -6 2

           -4 1 -4  1

          -1  8  0  -2

处理后变成：

            0 -2 -9 -9

            9 11 5  7 

            -4 -3 -7 -6

             -1  7  7  5

三重for循环，sum+=map[k][j]-map[k][i-1]。 ( map[k][j]-map[k][i-1]表示第k行i列和j列之间的元素和）

注意k为最内层循环，每次都是确定两列后，将行边累加，边找maxsum。

列数的确定也要按一定的顺序，首先i固定在1，j从i到n。然后依次增大i...其实就是外面的两重循环。

这样就能找到遍历所有的子矩阵，找到最大的sum.

注意：max不能初始化为0，因为最大sum可能为负。

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#define INF  0x3f3f3f3f
using namespace std;

int map[101][101];

int main()
{
    int tmp,i,j,k,n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(map,0,sizeof(map));
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                scanf("%d",&tmp);
                map[i][j]=map[i][j-1]+tmp;
            }
        }
        int maxsum=-INF,sum;
        for(i=1;i<=n;i++)
        {
            for(j=i;j<=n;j++)
            {
                sum=0;
                for(k=1;k<=n;k++)
                {
                    sum+=map[k][j]-map[k][i-1];
                    //printf("yes:  %d\n",map[k][j]-map[k][i-1]);
                    //printf("test: %d\n",sum);
                    if(sum<0)
                        sum=0;
                    if(sum>maxsum)
                        maxsum=sum;
                    //printf("test: %d %d\n",sum,maxsum);
                }
            }
        }
        printf("%d\n",maxsum);
    }
    return 0;
}