UVa 10003 - Cutting Sticks (White Book, Interval DP)

题目地址：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=12&problem=944&mosmsg=Submission+received+with+ID+17137424

思路：relatively naked rangeDP吧,状态转移方程为dp[i][j]=min(dp[i][k]+dp[k][j])+a[j]-a[i],The optimization can be done with quadrilateral inequalities

AC代码1：

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       #include <iostream>
       
       

       
       #include <cstdio>
       
       

       
       #include <cstdlib>
       
       

       
       #include <algorithm>
       
       

       
       #include <queue>
       
       

       
       #include <stack>
       
       

       
       #include <map>
       
       

       
       #include <cstring>
       
       

       
       #include <climits>
       
       

       
       #include <cmath>
       
       

       
       #include <cctype>
       
       

       
       const 
       
       int 
       
       inf 
       
       = 
       
       0x3f3f3f3f;
       
       //1061109567
       
       

       
       typedef 
       
       long 
       
       long 
       
       LL;
       
       

       
       #define lson l,m,rt<<1
       
       

       
       #define rson m+1,r,rt<<1|1
       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       int 
       
       a[
       
       55],
       
       dp[
       
       55][
       
       55];
       
       

       
       int 
       
       main()
       
       
{
       
       
    
       
       int 
       
       l;
       
       
    
       
       while(
       
       scanf(
       
       "%d",
       
       &
       
       l) 
       
       && 
       
       l)
       
       
    {
       
       
        
       
       memset(
       
       dp,
       
       inf,
       
       sizeof(
       
       dp));
       
       
        
       
       int 
       
       n;
       
       
        
       
       scanf(
       
       "%d",
       
       &
       
       n);
       
       
        
       
       a[
       
       0] 
       
       = 
       
       0;
       
       
        
       
       for(
       
       int 
       
       i
       
       =
       
       1; 
       
       i
       
       <=
       
       n; 
       
       i
       
       ++)
       
       
            
       
       scanf(
       
       "%d",
       
       &
       
       a[
       
       i]);
       
       
        
       
       a[
       
       n
       
       +
       
       1] 
       
       = 
       
       l;
       
       
        
       
       for(
       
       int 
       
       i
       
       =
       
       0; 
       
       i
       
       <=
       
       n; 
       
       i
       
       ++)
       
       
        {
       
       
            
       
       dp[
       
       i][
       
       i
       
       +
       
       1] 
       
       = 
       
       0;
       
       
        }
       
       
        
       
       for(
       
       int 
       
       r
       
       =
       
       3; 
       
       r
       
       <=
       
       n
       
       +
       
       2; 
       
       r
       
       ++)
       
       //注意这里的长度是n+2
       
       
        {
       
       
            
       
       for(
       
       int 
       
       i
       
       =
       
       0; 
       
       i
       
       <=
       
       n
       
       -
       
       r
       
       +
       
       2; 
       
       i
       
       ++)
       
       //注意这里i的取值范围
       
       
            {
       
       
                
       
       int 
       
       j 
       
       = 
       
       i 
       
       + 
       
       r 
       
       - 
       
       1;
       
       
                
       
       for(
       
       int 
       
       k
       
       =
       
       i
       
       +
       
       1; 
       
       k
       
       <
       
       j; 
       
       k
       
       ++)
       
       
                {
       
       
                    
       
       if(
       
       dp[
       
       i][
       
       j] 
       
       > 
       
       dp[
       
       i][
       
       k] 
       
       + 
       
       dp[
       
       k][
       
       j])
       
       
                        
       
       dp[
       
       i][
       
       j] 
       
       = 
       
       dp[
       
       i][
       
       k] 
       
       + 
       
       dp[
       
       k][
       
       j];
       
       
                }
       
       
                
       
       dp[
       
       i][
       
       j] 
       
       += 
       
       a[
       
       j] 
       
       - 
       
       a[
       
       i];
       
       
                
       
       //printf("%d %d %d\n",i,j,dp[i][j]);
       
       
            }
       
       
        }
       
       
        
       
       printf(
       
       "The minimum cutting is %d.\n",
       
       dp[
       
       0][
       
       n
       
       +
       
       1]);
       
       
    }
       
       
    
       
       return 
       
       0;
       
       
}
      
      
      
      

       
       
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AC代码2：

登录后复制

      
      

       
       #include <iostream>
       
       

       
       #include <cstdio>
       
       

       
       #include <cstdlib>
       
       

       
       #include <algorithm>
       
       

       
       #include <queue>
       
       

       
       #include <stack>
       
       

       
       #include <map>
       
       

       
       #include <cstring>
       
       

       
       #include <climits>
       
       

       
       #include <cmath>
       
       

       
       #include <cctype>
       
       

       
       const 
       
       int 
       
       inf 
       
       = 
       
       0x3f3f3f3f;
       
       //1061109567
       
       

       
       typedef 
       
       long 
       
       long 
       
       LL;
       
       

       
       #define lson l,m,rt<<1
       
       

       
       #define rson m+1,r,rt<<1|1
       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       int 
       
       a[
       
       55],
       
       dp[
       
       55][
       
       55];
       
       

       
       int 
       
       main()
       
       
{
       
       
    
       
       int 
       
       l;
       
       
    
       
       while(
       
       scanf(
       
       "%d",
       
       &
       
       l) 
       
       && 
       
       l)
       
       
    {
       
       
        
       
       memset(
       
       dp,
       
       inf,
       
       sizeof(
       
       dp));
       
       
        
       
       int 
       
       n;
       
       
        
       
       scanf(
       
       "%d",
       
       &
       
       n);
       
       
        
       
       a[
       
       0] 
       
       = 
       
       0;
       
       
        
       
       for(
       
       int 
       
       i
       
       =
       
       1; 
       
       i
       
       <=
       
       n; 
       
       i
       
       ++)
       
       
            
       
       scanf(
       
       "%d",
       
       &
       
       a[
       
       i]);
       
       
        
       
       a[
       
       n
       
       +
       
       1] 
       
       = 
       
       l;
       
       
        
       
       for(
       
       int 
       
       i
       
       =
       
       0; 
       
       i
       
       <=
       
       n; 
       
       i
       
       ++)
       
       
        {
       
       
            
       
       dp[
       
       i][
       
       i
       
       +
       
       1] 
       
       = 
       
       0;
       
       
        }
       
       
        
       
       for(
       
       int 
       
       r
       
       =
       
       3; 
       
       r
       
       <=
       
       n
       
       +
       
       2; 
       
       r
       
       ++)
       
       
        {
       
       
            
       
       for(
       
       int 
       
       i
       
       =
       
       0; 
       
       i
       
       <=
       
       n
       
       -
       
       r
       
       +
       
       3; 
       
       i
       
       ++)
       
       //One more time than the previous one,其实是没有必要的
       
       
            {
       
       
                
       
       int 
       
       j 
       
       = 
       
       i 
       
       + 
       
       r 
       
       - 
       
       1;
       
       
                
       
       for(
       
       int 
       
       k
       
       =
       
       i
       
       +
       
       1; 
       
       k
       
       <
       
       j; 
       
       k
       
       ++)
       
       
                {
       
       
                    
       
       if(
       
       dp[
       
       i][
       
       j] 
       
       > 
       
       dp[
       
       i][
       
       k] 
       
       + 
       
       dp[
       
       k][
       
       j])
       
       
                        
       
       dp[
       
       i][
       
       j] 
       
       = 
       
       dp[
       
       i][
       
       k] 
       
       + 
       
       dp[
       
       k][
       
       j];
       
       
                }
       
       
                
       
       dp[
       
       i][
       
       j] 
       
       += 
       
       a[
       
       j] 
       
       - 
       
       a[
       
       i];
       
       
            }
       
       
        }
       
       
        
       
       printf(
       
       "The minimum cutting is %d.\n",
       
       dp[
       
       0][
       
       n
       
       +
       
       1]);
       
       //注意这里是dp[0][n+1]
       
       
    }
       
       
    
       
       return 
       
       0;
       
       
}
      
      
      
      

       
       
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