Optimal Binary Search Tree - UVa 10304

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本文探讨了如何根据一组元素及其查询频率构建最优二叉搜索树的问题。文章提供了详细的输入输出示例，并附带了一段AC代码实现，该代码通过动态规划的方法计算出最优二叉搜索树的总成本。

Problem E

Optimal Binary Search Tree

Input: standard input

Output: standard output

Time Limit: 30 seconds

Memory Limit: 32 MB

Given a set S = (e₁, e₂, ..., e_n) of n distinct elements such that e₁ < e₂ < ... < e_n and considering a binary search tree (see the previous problem) of the elements of S, it is desired that higher the query frequency of an element, closer will it be to the root.

The cost of accessing an element e_i of S in a tree (cost(e_i)) is equal to the number of edges in the path that connects the root with the node that contains the element. Given the query frequencies of the elements of S, (f(e₁), f(e₂, ..., f(e_n)), we say that the total cost of a tree is the following summation:

f(e₁)*cost(e₁) + f(e₂)*cost(e₂) + ... + f(e_n)*cost(e_n)

In this manner, the tree with the lowest total cost is the one with the best representation for searching elements of S. Because of this, it is called the Optimal Binary Search Tree.

Input

The input will contain several instances, one per line.

Each line will start with a number 1 <= n <= 250, indicating the size of S. Following n, in the same line, there will be n non-negative integers representing the query frequencies of the elements of S: f(e₁), f(e₂), ..., f(e_n). 0 <= f(e_i) <= 100. Input is terminated by end of file.

Output

For each instance of the input, you must print a line in the output with the total cost of the Optimal Binary Search Tree.

Sample Input

1 5

3 10 10 10

3 5 10 20

Sample Output

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AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int val[260],sum[260],dp[260][260],ret;
int main()
{ int n,m,i,j,k,l,r,d;
  while(~scanf("%d",&n))
  { for(i=1;i<=n;i++)
    { scanf("%d",&val[i]);
      sum[i]=sum[i-1]+val[i];
    }
    for(d=2;d<=n;d++)
     for(l=1;l+d-1<=n;l++)
     { r=l+d-1;
       ret=1000000000;
       for(k=l;k<=r;k++)
        ret=min(ret,dp[l][k-1]+dp[k+1][r]+sum[r]-sum[l-1]-val[k]);
       dp[l][r]=ret;
     }
    printf("%d\n",dp[1][n]);
  }
}