uva 839

  Not so Mobile 

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

#include <iostream>
#include <stack>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <fstream>
#include <stack>
#include <list>
#include <sstream>
#include <cmath>

using namespace std;

#define ms(arr, val) memset(arr, val, sizeof(arr))
#define mc(dest, src) memcpy(dest, src, sizeof(src))
#define N 200
#define INF 0x3fffffff
#define vint vector<int>
#define setint set<int>
#define mint map<int, int>
#define lint list<int>
#define sch stack<char>
#define qch queue<char>
#define sint stack<int>
#define qint queue<int>
bool ans;
int n;

int build()
{
    int wl, dl, wr, dr;
    scanf("%d%d%d%d", &wl, &dl, &wr, &dr);
    if(!wl) 
        wl = build();//zuo
    if (!wr)
        wr = build();//you
    if (wl * dl != wr * dr)
        ans = false;
    return wl + wr;
}

void input()
{
    if (ans)
    {
        puts("YES");
    }
    else
    {
        puts("NO");
    }
    if (n)
    {
        putchar('
');
    }
}

int main()
{
    scanf("%d", &n);
    while (n--)
    {
        ans = true;
        build();
        input();
    }
    return 0;
}