Uva

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

递归的输入计算，想清楚递归的形式还是很好做的。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <string>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <stack>
#include <queue>
#include <bitset> 
#include <cassert> 

using namespace std;

// 递归的输入并判断是否平衡。
bool solve(int& W)
{
	int Wl, Dl, Wr, Dr;
	cin >> Wl >> Dl >> Wr >> Dr;
	bool b1 = true, b2 = true;
	if (!Wl) { // 左边是个子天平，递归输入求解
		b1 = solve(Wl);
	}
	if (!Wr) { // 右边是个子天平，递归输入求解
		b2 = solve(Wr);
	}
	W = Wl + Wr; // 左右都计算完毕，修改W为子天平的总重量
	return b1 && b2 && (Wl * Dl == Wr * Dr);
}

int main()
{
	ios::sync_with_stdio(false);
	int T, W;
	cin >> T;
	while (T--) {
		if (solve(W)) {
			cout << "YES
";
		}
		else {
			cout << "NO
";
		}
		if (T) {
			cout << endl;
		}
	}

	return 0;
}