UVA839——天平

描述：

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.



分析：

简而言之，输入一个天平系统，从上到下输入，如果该位置有子天平，则按照由左到右的顺序输入，判断该系统是否平衡（即左重量*左力臂=右重量*右力臂）

该题输入采用了递归定义的方式，因此题解用递归写也很方便，此处使用传值引用，代码可以写的很简洁。

P.S该题提交的时候遇到了点小问题，到现在没搞明白，疑点已经写在注释里了，求高人赐教。

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>

//#define file
#define PI 3.1415926
#define MAX(A,B) ((A)>(B)?(A):(B))
#define ll long long
#define fordo(A,B,C) for(int (A)=(B);(A)<=(C);(A)++)
using namespace std;

bool solve(int& W)
{
int W1,D1,W2,D2;
bool b1=true;
bool b2=true;
cin>>W1>>D1>>W2>>D2;
if(!W1)//这里很奇怪，当我用W1==0时VJ就会报WA，只有！W1这种表示法可以过，貌似这两者没区别吧……
b1=solve(W1);//当存在子天平，求解。
if(!W2)
b2=solve(W2);
W=W1+W2;//由子天平重量计算该天平重量
return b1&&b2&&(W1*D1==W2*D2);
//递归求解：当该托盘的子天平都平衡，并且该天平平衡，则返回该天平平衡。
}

int main()
{
#ifdef file
freopen("test.in", "r", stdin);
freopen("test.out", "w", stdout);
#endif // file
int T,W;
cin>>T;
while(T--)
{
if(solve(W))
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
if(T)
cout<<endl;
}
return 0;
}