P1103 Book sorting and p1103 Book sorting

P1103 Book sorting and p1103 Book sorting
Description

Frank is a neat person. He has a lot of books and a bookshelf and wants to put the books on the shelf. The bookshelves can put down all the books, so Frank first arranged the books in High Order on the bookshelves. However, Frank found that because many books have different widths, the books still look very untidy. So he decided to remove k books from the book so that the bookshelves could look neat.

The non-uniformity of the bookshelves is defined as the sum of the absolute values of the difference in the width of each two books. For example, there are four books:

1x2 5x3 2x4 3x1 then Frank arranged it neatly as follows:

1x2 2x4 3x1 5x3 untidy is 2 + 3 + 2 = 7

We know that the height of each book is different. Please find the minimum untidy degree after removing k books.

Input/Output Format

Input Format:

 

The numbers n and k in the first line represent the number of books and the number of books removed from them. (1 <= n <= 100, 1 <= k <n)

In the n rows below, two numbers in each row indicate the height and width of a book, both smaller than 200.

Ensure high repetition

 

Output Format:

 

An integer in one row indicates the minimum untidy degree of the bookshelf.

 

Input and Output sample input sample #1:

4 11 22 43 15 3

Output sample #1:

3

We use reverse thinking.
Use dp [I] [j] to represent the previous I book, leaving the minimum value of j book

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #include<algorithm> 7 using namespace std; 8 const int MAXN=301; 9 void read(int &n)10 {11     char c='+';int x=0;bool flag=0;12     while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}13     while(c>='0'&&c<='9'){x=x*10+c-48;c=getchar();}14     flag==1?n=-x:n=x;15 }16 int n,k;17 struct node18 {19     int gao,kuan;20 }book[MAXN];21 int comp(const node &a,const node &b)22 {23     return a.gao<b.gao;24 }25 int dp[MAXN][MAXN];26 int presum[MAXN];27 int main()28 {29     read(n);read(k);30     k=n-k;31     for(int i=1;i<=n;i++)32     {33         read(book[i].gao);34         read(book[i].kuan);35     }36     sort(book+1,book+n+1,comp);37     for(int i=2;i<=n;i++)38         for(int j=2;j<=min(k,i);j++)39         {40             dp[i][j]=0x7fffff;41             for(int k=j-1;k<i;k++)42                 dp[i][j]=min(dp[i][j],dp[k][j-1]+abs(book[i].kuan-book[k].kuan));43         }44     int ans=0x7fffff;45     for(int i=k;i<=n;i++)46         ans=min(ans,dp[i][k]);47     printf("%d",ans);48     return 0;49 }