UVA-108-Maximum Sum-子矩阵最大和（最大连续子序列的变形）+降维处理+dp

A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem. 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 subrectangle 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:   　　9    2

　　　　　　　　　　　　　　　　 −4    1

　　　　　　　　　　　　　　　　−1     8         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 N2 integers separated by white-space (newlines and spaces). These N2 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

0      −2     −7     0

9       2      −6     2

−4     1      −4      1

−1     8       0     −2

Sample Output

15

题意：从所给的 N × N 的矩阵中选出任意矩阵，使其中的元素和最大

思路：

　　枚举i、j，然后对j进行降维压缩，再对降维后的一维数组求最大子序列（利用动态规划）即可。

　　（还看到一种写法，但是不是特别理解：枚举行区间，求出每列的和，再用d[i]=max(d[i-1]+sum[i],sum[i])动态规划公式，即可求出最大子矩阵和。）

降维压缩代码：

 #include<iostream>
 #include<string.h>
 #include<algorithm>
 #define inf 0x3f3f3f3f
 using namespace std;

 int a[][];
 int c[];
 int dp[];
 int main()
 {
     std::ios::sync_with_stdio(false);
     int n;
     cin>>n;
     for(int i=; i<n; i++)
     {
         for(int j=; j<n; j++)
         {
             cin>>a[i][j];
         }
     }
     int ans=-inf;
     for (int i=; i<n; i++) //n行
     {
         memset(c,,sizeof(c));
         for (int j=i; j<n; j++) //n行
         {
             //对i到j行矩阵进行降维操作
             for (int k=; k<n; k++) //n列
             {
                 c[k]+=a[j][k];
             }
             dp[]=c[];//对降维后的数组c[k]进行最大子序列和的动态规划
             if (ans<dp[])
             {
                 ans = dp[];
             }
             for(int k=;k<n;k++)// n列
             {
                 dp[k]=max(dp[k-]+c[k],c[k]);
                 if (ans<dp[k])
                 {
                     ans = dp[k];
                 }
             }
         }
     }
     cout<<ans<<endl;
     return ;
 }

不是特别理解的代码：

枚举行区间，求出每列的和，再用d[i]=max(d[i-1]+sum[i],sum[i])动态规划公式，即可求出最大子矩阵和。

 #include<iostream>
 #define inf 0x3f3f3f3f
 using namespace std;

 int a[][];
 int main()
 {
     std::ios::sync_with_stdio(false);
     int n;
     cin>>n;
     for(int i=; i<=n; i++)
     {
         for(int j=; j<=n; j++)
         {
             cin>>a[i][j];
             a[i][j]+=a[i-][j];
         }
     }//每一行等于加上上面一行的和
     int maxx=a[][];
     for(int i=; i<=n; i++)
     {
         for(int j=i+; j<=n; j++)
         {
             int sum=;
             for(int k=;k<=n;k++)
             {
                 sum+=a[j][k]-a[i][k];
                 if(sum<)
                     sum=;
                 if(sum>maxx)
                     maxx=sum;
             }
         }
     }
     cout<<maxx<<endl;

     return ;
 }