UVA10304 Optimal Binary Search Tree最优排序二叉树

Posted 2021-06-29 海岛Blog

Given a set S = (e1, e2, …, en) of n distinct elements such that e1 < e2 < . . . < en 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 ei of S in a tree (cost(ei)) 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(en)), we say that the total cost of a tree is the following summation:
f(e1) ∗ cost(e1) + f(e2) ∗ cost(e2) + . . . + f(en) ∗ cost(en)
    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(en), 0 ≤ f(ei) ≤ 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
0
20
20

问题链接：UVA10304 Optimal Binary Search Tree
问题简述：（略）
问题分析：最优排序二叉树（OBST）问题，不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA10304 Optimal Binary Search Tree */

#include <bits/stdc++.h>

using namespace std;

const int INF = 0x3F3F3F3F;
const int N = 250 + 1;
int a[N], sum[N], dp[N][N];

int main()
{
    int n;
    while (scanf("%d", &n) == 1) {
        sum[0] = 0;
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            sum[i] = sum[i - 1] + a[i];
        }

        memset(dp, 0, sizeof dp);
        for (int d = 2; d <= n; d++)
            for (int l = 1; l <= n - d + 1; l++) {
                int r = l + d - 1;
                int tot = sum[r] - sum[l - 1];
                int ans = INF;
                for (int k = l; k <= r; k++)
                    ans = min(ans, dp[l][k - 1] + dp[k + 1][r] + tot - a[k]);
                dp[l][r] = ans;
            }

        printf("%d\\n", dp[1][n]);
    }

    return 0;
}