hdu 1081最大子矩阵的和DP

To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8210 Accepted Submission(s): 3991



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这道题题意很清楚：给一个N*N的矩阵，求它的子矩阵（元素和最大）；（1）首先得清楚 对于一维数组如何求它的最大子段和；    
int sum=0,Max=-9999999;
for(int i=1;i<=n;i++)//a数组一共用n个元素，遍历一遍即可
{
 if(sum<0)sum=0; //如果sum都小于0了,前面的就可以不要了，从头再来，初始化0
 sum+=a[i];
 if(sum>Max)Max=sum;   //不断更新Max,最终的Max就是最大字段和
}
（2）清楚一维最大字段和的解法之后，再接着看二维的，经典方法就是将二维的压缩为一维的，如果这个压缩过程想明白了就简单了；举个例子：4*4的矩阵 0 -2 -7 0 9  2 -6 2-4  1 -4 1-1  8  0 -2 1.将第一列看成一维的求出最大的Max; 2.将第一列和第二列合并 看成一维的求出最大的Max 3.将第一列，第二列，第三列合并，看成一维的求出最大的Max; 4.将第一列，第二列，第三列，第四列合并，看成一维的求出最大的Max; 
 5.将第二列看成一维 求最值 6.将第二列，第三列合并成一维的求最值 7.将第二列，第三列，第四列合并一维 求最值
 8.将第三列看成一维的，求最值 9.将第三列，第四合并一维求最值； 
 10.将第四列看成一维的，求最值End  ~~~~~~压缩过程就是这样的~~应该能看懂！！！  那么  压缩之前稍微处理一下：  dp[i][j]表示 第i行前j列元素之和;那么比如合并 第二列和第三列的时候就是这样 ：  dp[1][3]-dp[1][2-1] 就是说第一行的前3个元素的和减去第一行的前1个元素的和当作一维的处理~~~依次类推[code]

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
const int maxn=110;
using namespace std;
int dp[maxn][maxn];
int main()
{
 int n;
 while(cin>>n)
 {
  memset(dp,0,sizeof(dp));
  for(int i=1;i<=n;i++)  //dp[i][j]表示第i行的前j个元素
   for(int j=1;j<=n;j++)
   {
    cin>>dp[i][j];
    dp[i][j]+=dp[i][j-1];
   }

  int Max=-99999999;
  for(int i=1;i<=n;i++)
   for(int j=i;j<=n;j++)
   {
     int sum=0;
     for(int k=1;k<=n;k++)
     {
       if(sum<0)sum=0;
       sum+=dp[k][j]-dp[k][i-1];  //dp[k][j]-dp[k][i-1]表示第k行第i到第j个元素的和
       if(sum>Max)Max=sum;
     }
   }
   cout<<Max<<endl;
 }
 return 0;
}

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