G

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-8;
const int mod=100000007;
const int MS=10005;

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 0;
}