ural 1146. Maximum Sum（动态规划）

1146. Maximum Sum

Time limit: 1.0 second Memory limit: 64 MB

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 sub-rectangle 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 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 Non a line by itself indicating the size of the square two dimensional array. This is followed by N ²integers separated by white-space (newlines and spaces). These N ² 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	output
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2	15

题意：输入部分第一行是一个正整数N,说明矩阵的大小。下一行后面跟着 个整数。留意这些整数的输入格式可能比较混乱。矩阵是以行优先存储的。N可能和100一样大，矩阵中每个数字的范围为 [-127, 127].

思路1:对矩阵进行穷举。找出所有的子矩阵计算矩阵和并找出最大值。

 #include<iostream>
 #include<cstdio>
 #include<string>
 #include<cstring>

 using namespace std;
 int s[][]= {};
 int dp[][]= {};

 int main()
 {
 //    freopen("1.txt","r",stdin);
     int i,j,ki,kj;
     int n;
     while(cin>>n)
     {
         for(i=; i<=n; i++)
             for(j=; j<=n; j++)
             {
                 cin>>s[i][j];
                 s[i][j]+=s[i-][j];
                 s[i-][j]+=s[i-][j-];
             }//输入矩阵并且计算从（1,1）到（i-1，j）的矩阵的和值
         for(i=; i<=n; i++)
             s[n][i]+=s[n][i-];//计算从（1,1）到（n，i）的矩阵的和值
         int max=-;//特别注意因为
         for(i=; i<=n; i++)
             for(j=; j<=n; j++)
             {
                 max=-;
                 for(ki=; ki<i; ki++)
                     for(kj=; kj<j; kj++)
                         if(max<s[i][j]-s[i][kj]-s[ki][j]+s[ki][kj])//计算（ki，kj）到（i,j)的子矩阵的和并且和max比较；
                             max=s[i][j]-s[i][kj]-s[ki][j]+s[ki][kj];
                 dp[i][j]=max;
             }
         max=-;
         for(i=; i<=n; i++)
             for(j=; j<=n; j++)
                 if(max<=dp[i][j])
                     max=dp[i][j];
         cout<<max<<endl;
     }
     return ;
 }

思路2求最大子矩阵的和，先按列进行枚举起始列和结束列，用数据够快速求出每行起始列到结束列之间各数据的和，然后就可以降成一维的，问题就变成了求最大连续子序列了

 #include<cstdio>
 #include<cstring>
 #include<iostream>

 using namespace std;

 #define N 1100
 #define INF 0x3f3f3f3f
 int s[N][N]= {},dp[N],n;

 void get_sum(int x ,int y)
 {
     for(int i=; i<=n; i++)
     {
         dp[i]=s[y][i]-s[x-][i];
     }
     return ;
 }

 int DP()
 {
     int sum=,max=-INF;
     for(int i=; i<=n; i++)
     {
         sum+=dp[i];
         max=sum>max?sum:max;
         if(sum<) sum=;
     }
     return max;
 }

 int main()
 {
     int i,j;
 //    freopen("1.txt","r",stdin);
     while(cin>>n)
     {
         for(i=; i<=n; i++)
             for(j=; j<=n; j++)
             {
                 cin>>s[i][j];
                 s[i][j]+=s[i-][j];
             }//输入矩阵并且计算从（1,j）到（i，j）的矩阵的和值
         int max=-INF,ans;
         for(i=; i<=n; i++)
             for(j=i; j<=n; j++)
             {
                 get_sum(i,j);
                 ans=DP();
                 max=ans>max?ans:max;
             }
         printf("%d\n",max);
     }
     return ;
 }