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Not so Mobile

2024-08-03 16:57:51   225人阅读

Not so Mobile

Before being an ubiquous communications gadget, amobile 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.

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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 isWl×Dl = Wr×Dr whereDl is the left distance, Dr is the right distance, Wl is the left weight andWr 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.

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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 WrDr

If Wl orWr 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 andWr 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

二叉树递归：先给出根结点的左右结点，w1 = 0,d1 = 2,w2 = 0,d2 = 4;那么左节点为零则读当前左节点的左右孩子的值；知道左右孩子都有值为止；
# include <iostream>
# include <cstdio>
using namespace std;

int flag;

int read()
{
    int w1,d1,w2,d2;
    cin>>w1>>d1>>w2>>d2;
    if(w1&&w2&&d1&&d2)    //如果左右孩子都存在，那么则进行题目中的判断；
    {
        if(w1*d1!=w2*d2)    {flag=1;   return 0;}
        else return (w1+w2);    //返回根结点的值；
    }
    else
    {
        if(!w1)    w1=read();   //结点的值为零时继续读数据;
        if(!w2)    w2=read();
        if(w1*d1!=w2*d2)    {flag=1;   return 0;}
        else return (w1+w2);
    }
}

int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        flag=0;
        read();   //读取数据;
        if(!flag)    cout<<"YES"<<endl;
        else   cout<<"NO"<<endl;
        if(n!=0)    cout<<endl;
    }
    return 0;
}


Not so Mobile


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