Optimal Binary Search Tree+区间dp+uva_given a set s = (e1, e2, ..., en) of n distinct el-优快云博客

本文探讨了如何通过区间动态规划方法实现最优二叉搜索树的构建，优化查询元素的路径成本，降低总查找成本。

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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_iof 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

解决方案：此题为区间dp，选择根节点k,dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j]+sum[k-1]-sum[i-1]+sum[j]-sum[k])，sum为数组前n项和。每个节点到根节点的cost可用区间叠加来代替。

code：#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int dp[550][550];
int num[550];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(num,0,sizeof(num));
        for(int i=1; i<=n; i++)
        {
            int temp;
            scanf("%d",&temp);
            num[i]=num[i-1]+temp;
          //  cout<<num[i]<<endl;
        }
        memset(dp,0,sizeof(dp));
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j+i<=n; j++)
            {
                dp[j][j+i]=0x3f3f3f3f;
                for(int k=j; k<=j+i; k++)
                {
                    int temp=0;
                    if(k!=j){
                        temp+=dp[j][k-1]+num[k-1]-num[j-1];
                    }
                    if(k!=i+j){
                        temp+=dp[k+1][j+i]+num[j+i]-num[k];
                    }
                    dp[j][j+i]=min(dp[j][j+i],temp);
                }
            }
        }
        printf("%d\n",dp[1][n]);
    }
    return 0;
}