G - Not so Mobile

G - Not so Mobile

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

 

 

Description

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 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.

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

1

0 2 0 4
0 3 0 1
1 1 1 1
2 4 4 2
1 6 3 2

Sample Output

　　YES

一道巧妙使用引用巧妙解答的题目。有点小思维难度。是一道不错的题目。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <stack>
 #include <queue>
 #include <set>
 #include <map>
 #include <list>
 #include <iomanip>
 #include <cstdlib>
 #include <sstream>
 using namespace std;
 const int INF=0x5fffffff;
 const double EXP=1e-;
 const int mod=;
 const int MS=;

 bool solve(int &w)
 {
     int w1,d1,w2,d2;
     bool b1=true,b2=true;
     cin>>w1>>d1>>w2>>d2;
     if(!w1)
         b1=solve(w1);
     if(!w2)
         b2=solve(w2);
     w=w1+w2;
     return b1&&b2&&(w1*d1==w2*d2);
 }

 int main()
 {
     int T,W;
     cin>>T;
     while(T--)
     {
         if(solve(W))
             cout<<"YES"<<endl;
         else
             cout<<"NO"<<endl;
         if(T)
             cout<<endl;
     }
     return ;
 }