G - Not so Mobile

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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。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <string>
 7 #include <vector>
 8 #include <stack>
 9 #include <queue>
10 #include <set>
11 #include <map>
12 #include <list>
13 #include <iomanip>
14 #include <cstdlib>
15 #include <sstream>
16 using namespace std;
17 const int INF=0x5fffffff;
18 const double EXP=1e-8;
19 const int mod=100000007;
20 const int MS=10005;
21 
22 bool solve(int &w)
23 {
24     int w1,d1,w2,d2;
25     bool b1=true,b2=true;
26     cin>>w1>>d1>>w2>>d2;
27     if(!w1)
28         b1=solve(w1);
29     if(!w2)
30         b2=solve(w2);
31     w=w1+w2;
32     return b1&&b2&&(w1*d1==w2*d2);
33 }
34 
35 int main()
36 {
37     int T,W;
38     cin>>T;
39     while(T--)
40     {
41         if(solve(W))
42             cout<<"YES"<<endl;
43         else
44             cout<<"NO"<<endl;
45         if(T)
46             cout<<endl;
47     }
48     return 0;
49 }

G - Not so Mobile

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原文地址：http://www.cnblogs.com/767355675hutaishi/p/4321623.html