839 - Not so Mobile

最新推荐文章于 2024-08-28 21:51:32 发布

SIO__Five

最新推荐文章于 2024-08-28 21:51:32 发布

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分类专栏： UVA 文章标签：算法 UVa

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Not so Mobile

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.

$\epsfbox{p839a.eps}$

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 W_l×D_l = W_r×D_r where D_l is the left distance, D_r is the right distance, W_l is the left weight and W_r 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.

$\epsfbox{p839b.eps}$

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: W_l D_l W_r D_r

If W_l or W_r 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 W_l and W_r 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

Sample Output

YES

由于二叉树的建立可以用递归来实现。所以和二叉树相关的题目，都能够读出递归的味道。

这道题是一个系统，要求能够达到杠杆平衡。每一组为一个单位，那其左侧和右侧有可以再次分为两个小的平衡体来做。

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
using namespace std;
bool uq=true;
int input()
{
    int w1,d1,w2,d2;
    cin>>w1>>d1>>w2>>d2;
    if (w1==0)
        w1+=input();
    if (w2==0)
        w2+=input();
    if (w1*d1!=w2*d2) uq=false;
    return w1+w2;
}
int main ()
{
    int n;
    cin>>n;
    while(n--)
    {
        uq=true;
        input();
        if (uq) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
        if (n!=0) cout<<endl;
    }
}