poj1050(nyoj104 zoj1074)dp问题

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 39913		Accepted: 21099

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*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 * 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

这是hdu 1003的拓展，知道了在一维数组中如何求最大连续子段和，那么这题就是扩展到二维数组中，让我们求出子矩阵最大的和，我们可以这样考虑，我们把同行不同列（或者同列不同行）的加起来，比如i行，j行，i,j两行之间的数字组成了一个矩阵，我们把i行到j行之间同列的数组元素加起来按照列号组成一个一维数组，这样我们只需要利用最大子段和的知识找出这个数组的最大连续和，这个和就是我们要求的那个子矩阵最大和！ 可以说， i和j行定义了子矩阵高度，一维数组最大子段和连续的长度定义了子矩阵的宽度，OK，代码。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;

 int temp[],n;
 int cal()  //
 {
     int Max = temp[];
     int sum = temp[];
     for(int i = ;i<n;i++)
     {
         if(sum+temp[i]<temp[i])
             sum = temp[i];
         else
             sum+=temp[i];
         if(Max<sum)
             Max = sum;
     }
     return Max;
 }

 int main()
 {
     while(scanf("%d",&n)!=EOF)
     {
         int a[][];
         for(int i = ;i<n;i++)
             for(int j = ;j<n;j++)
                 scanf("%d",&a[i][j]);
         int Max = ;
         for(int i = ;i< n;i++)  //i是起始行
         {
             for(int j =i ;j<n;j++)    //j是终止行
             {
                 memset(temp,,sizeof(temp));
                 for(int m = ;m<n;m++)   //固定列,注意是行在变
                 {
                     for(int k =i ;k<=j;k++)  //累加i起始行，j终止行中间的同列的数据
                         temp[m]+=a[k][m];
                 }
                 int MaxTemp = cal();
                 if(MaxTemp>Max)
                     Max = MaxTemp;
             }

         }
         printf("%d\n",Max);
     }
     return ;
 }

事实证明我蠢了，后来看到nyoj这题的最优程序解答，在处理第i行到j行同列相加上面处理的很好，利用输入时候进行累加，然后做减法，直接减掉了一层循环

nyoj 的版本

 #include<iostream>
 #include<cstring>
 using namespace std;
 #define N 110
 int a[N][N];
 int b[N];
 int main()
 {
     int n,r,c;
     cin>>n;
     while(n--)
     {
         cin>>r>>c;
         for(int i=;i<=r;++i)
             for(int j=;j<=c;++j)
             {
                 cin>>a[i][j];
                 a[i][j]+=a[i-][j];
             }
         int max=a[][];
         for(int i=;i<=r-;++i)
             for(int j=i+;j<=r;++j)
             {
                 memset(b,,sizeof(b));
                 for(int k=;k<=c;++k)
                 {
                     if(b[k-]>=)
                         b[k]=b[k-]+a[j][k]-a[i][k];
                     else
                         b[k]=a[j][k]-a[i][k];
                     if(max<b[k])
                         max=b[k];
                 }
             }
         cout<<max<<endl;
     }
 }                        

还有一种没有用这种求和的方法，但是是先求第一行最大子段和，再求第一行跟第二行合起来的最大子段和，，再求第一到第三合起来的最大子段和，以此类推，直到求出整个矩阵的合起来的最大子段和，最后就是我们需要的解 。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;

 int temp[],n;
 int cal()  //最大子段和
 {
     int Max = temp[];
     int sum = temp[];
     for(int i = ;i<n;i++)
     {
         if(sum+temp[i]<temp[i])
             sum = temp[i];
         else
             sum+=temp[i];
         if(Max<sum)
             Max = sum;
     }
     return Max;
 }

 int main()
 {
     while(scanf("%d",&n)!=EOF)
     {
         int a[][];
         for(int i = ;i<n;i++)
             for(int j = ;j<n;j++)
                 scanf("%d",&a[i][j]);
         int Max = ;
         for(int i = ;i< n;i++)
         {
             memset(temp,,sizeof(temp));
             for(int j =i ;j<n;j++)
             {
                 for(int k = ;k<n;k++)
                     temp[k]+=a[j][k];
                 int t = cal();
                 if(t>Max)
                     Max =t;
             }

         }
         printf("%d\n",Max);
     }
     return ;
 }