Uva 839 Not so Mobile

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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
本题极为重要！！！

 1 #include<iostream>  
 2 #include<string.h>  
 3 #include<stdio.h>  
 4 #include<ctype.h>  
 5 #include<algorithm>  
 6 #include<stack>  
 7 #include<queue>  
 8 #include<set>  
 9 #include<math.h>  
10 #include<vector>  
11 #include<map>  
12 #include<deque>  
13 #include<list>  
14 using namespace std;  
15 bool solve(int& W)
16 {
17     int W1,D1,W2,D2;
18     bool b1=true,b2=true;
19     cin>>W1>>D1>>W2>>D2;
20     if(!W1)b1=solve(W1);
21     if(!W2)b2=solve(W2);
22     W=W1+W2;
23     return b1 && b2 && (W1*D1==W2*D2); 
24 } 
25 int main()
26 {
27     int T,W;
28     cin>>T;
29     while(T--)
30     {
31         if(solve(W))
32         cout<<"YES\n";
33         else
34         cout<<"NO\n";
35         if(T)
36         printf("\N");
37     }
38     return 0;
39 } 

View Code

Uva 839 Not so Mobile,布布扣,bubuko.com

Uva 839 Not so Mobile

标签：des   style   blog   http   color   strong   

原文地址：http://www.cnblogs.com/qscqesze/p/3853032.html