Not so Mobile (二叉搜索树的左右权值和相等)

博客围绕判断天平是否平衡展开,给出输入输出格式及示例。输入包含多组数据,需根据力矩原理判断。思路是在输入时构建树,根据输入特性回溯传递节点重量和进行比较。最后给出了实现该功能的代码思路。

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Before being an ubiquous communications gadget, a mobile was just a structure made of strings and wires suspending colourfull things. This kind of mobile is usually found hanging over cradles of small babies

The figure illustrates a simple mobile. It is just a wire, suspended by a string, with an object on each side. It can also be seen as a kind of lever with the fulcrum on the point where the string ties the wire. From the lever principle we know that to balance a simple mobile the product of the weight of the objects by their distance to the fulcrum must be equal. That is Wl × Dl = Wr × Dr where Dl is the left distance, Dr is the right distance, Wl is the left weight and Wr is the right weight.

In a more complex mobile the object may be replaced by a sub-mobile, as shown in the next figure. In this case it is not so straightforward to check if the mobile is balanced so we need you to write a program that, given a description of a mobile as input, checks whether the mobile is in equilibrium or not.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The input is composed of several lines, each containing 4 integers separated by a single space. The 4 integers represent the distances of each object to the fulcrum and their weights, in the format: Wl Dl Wr Dr

If Wl or Wr is zero then there is a sub-mobile hanging from that end and the following lines define the the sub-mobile. In this case we compute the weight of the sub-mobile as the sum of weights of all its objects, disregarding the weight of the wires and strings. If both Wl and Wr are zero then the following lines define two sub-mobiles: first the left then the right one.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

Write ‘YES’ if the mobile is in equilibrium, write ‘NO’ otherwise

Sample Input

1

0 2 0 4

0 3 0 1

1 1 1 1

2 4 4 2

1 6 3 2

Sample Output

YES

题意:t组数据,每组数据有若干小数据 代表天平的左右端 WL(左边端点重量,为0时表示有子树,继续输入子树)   DL(左天平臂长) WR(右边端点重量,为0时表示有子树,继续输入子树) DR (右天平臂长)。根据力矩(力 乘 力臂)判断是否是一个平衡的天平。

思路:由于是若干数据,不明确是多少,但数据都合法,那么我们就在输入数据的时候正好可以构建树,当输入数完成后就构建成了一颗完整的树(端点有重量表示那是叶节点)。然后我们可以根据输入的特性,即先根再左再右,如果遇到端点有重量时不再向下扩展,应向上回溯继续输入。在回溯的过程中正好传递此节点以下的重量和,并与另一端的装量和比较,再回溯。

代码如下:

#include<cstdio>
#include<cstring>
int sove(int &w)  //w 相当于此分支目前的重量
{
    int wl,dl,wr,dr;
    scanf("%d%d%d%d",&wl,&dl,&wr,&dr);
    int l=1,r=1; //如果到达页的时候最起码是可以比较
    if(!wl) //有左子树
        l=sove(wl);//把wl的地址传进去,可以在下一次递归的第十二行把重量值计算出来
    if(!wr) //有右子树
        r=sove(wr); //同理,获得此节点以下的重量和
    w=wl+wr;  //左右子树重量相加
    if(l&&r&&(wl*dl==wr*dr)) //左边,右边子树重量相等,再比较此时它的重量
        return 1;
    else
        return 0;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int x;
        int flag=sove(x);//x 相当于此分支目前的重量
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
        if(t)
            printf("\n");
    }
}

 

最优二叉搜索树(Optimal Binary Search Tree)问题是一个经典的动态规划问题。在计算权值时,我们需要考虑两个方面:节点的键值节点的概率。 首先,我们需要给定一组节点的键值它们对应的概率。假设这组节点的键值为 keys[1...n],对应的概率为 probabilities[1...n]。其中,keys[i] 表示第 i 个节点的键值,probabilities[i] 表示第 i 个节点的概率。 接下来,我们可以使用动态规划来计算最优二叉搜索树权值。 我们定义一个二维数组 dp[1...n+1][1...n],其中 dp[i][j] 表示子树 keys[i...j] 的最小权值。 首先,我们初始化只有一个节点的子树的权值。对于所有的 i,有 dp[i][i] = probabilities[i]。 然后,我们开始计算包含两个节点的子树的权值。对于所有的 i,有 dp[i][i+1] = probabilities[i] + probabilities[i+1]。 接下来,我们计算包含三个节点的子树的权值。对于所有的 i,有 dp[i][i+2] = probabilities[i] + probabilities[i+1] + probabilities[i+2]。 更一般地,对于包含 k 个节点的子树(k ≥ 4),我们可以使用以下递推公式计算权值: dp[i][j] = min(dp[i][j], dp[i][r-1] + dp[r+1][j] + sum(probabilities[i...j])), 其中,r 表示根节点的索引,i ≤ r ≤ j。 最后,dp[1][n] 就是整个最优二叉搜索树权值。可以通过迭代或递归的方式计算出最优权值。 需要注意的是,上述计算过程中,我们假设节点的键值是已经按照升序排列的。如果键值没有排序,我们可以先对键值进行排序,然后按照上述方法计算权值。 希望这个解答对你有帮助!如果你还有其他问题,请随时提问。
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