uva 839 - Not so Mobile

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 Wl×Dl = Wr×Dr where Dl is the left distance, Dr is the right distance, Wl is the left weight and Wr 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:
Wl Dl Wr Dr

If Wl or Wr 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 Wl and Wr 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('\n');
}
}

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