uva 839 Not so Mobile-S.B.S.

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 gure 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 gure.
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 dene
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 dene two sub-mobiles: rst 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

------------------------我是分割线----------------------------------------------------------------------------------------

这道题从题目就可以看出是递归关系定义的，所以使用递归进行输入；

并且可以在输入过程中进行判断；

使用引用传值而不用全局变量，极大简化代码，增加可读性。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<queue>
 #include<cmath>
 #include<algorithm>
 #include<cstdlib>
 using namespace std;
 int read(){
     int x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 bool build(int &w)
 {
     int w1,w2,d1,d2;
     bool b1=true,b2=true;
     w1=read();d1=read();w2=read();d2=read();
     if(!w1) b1=build(w1);
     if(!w2) b2=build(w2);
     w=w1+w2;
     return b1&&b2&&(w1*d1==w2*d2);
 }
 int main()
 {
     int t,w;
     t=read();
     while(t--)
     {
         if(build(w)) cout<<"YES"<<endl;
         else cout<<"NO"<<endl;
         if(t) cout<<"\n";
     }
     return ;
 }