天平(Not so Mobile)

Not so Mobile

Time Limit:3000MS Memory Limit:Unknown 64bit IO Format:%lld & %llu

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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 isW_l×D_l =W_r×D_r whereD_l is the left distance,D_r is the right distance, W_l is the left weight andW_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 bothW_l andW_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 

10 2 0 40 3 0 11 1 1 12 4 4 21 6 3 2

Sample Output 

YES

【分析】

       采用递归方式输入。

用C++语言编写程序，代码如下：

#include<iostream>using namespace std;//输入一个子天平，返回子天平是否平衡，参数w修改为子天平的总重量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;elsecout << "NO" << endl;if (T)cout << endl;}return 0;}

用java语言编写程序，代码如下：

（在测试过程中，曾在递归方法中加入一句输出语句，在提交过程中，忘记对该语句的删除，导致程序时间超限。删除该语句之后就答案正确了）

以下有两个方法可以使得在程序的递归方法中可以传递引用：

方法1：

import java.io.BufferedInputStream;import java.util.Scanner;//要注意的是java中并没有所谓的引用传递，当我们传递对象时，方法接受到的对象只不过是指向相同地址值的引用摆了，当我们改变该对象的指向时，并不会影响原来的地址值。//当我们只是改变对象本身，而没有改变对象的属性时，原对象的内容不变。（这也是为什么下面用Integer代替int改为传递对象不行的原因，Integer对象在方法中被//重新赋值，也不会影响调用者的Integer对象）public class Main {public static void main(String[] args) {Scanner input = new Scanner(new BufferedInputStream(System.in));int[] W = new int[1];int T = input.nextInt();for(int i = 0; i < T; i++) {if(i != 0)System.out.println();if(solve(input, W))System.out.println("YES");elseSystem.out.println("NO");}}public static boolean solve(Scanner input, int[] W) {int[] W1 = new int[1]; int[] W2 = new int[1];int D1,D2;boolean b1 = true, b2 = true;W1[0] = input.nextInt(); D1 = input.nextInt();W2[0] = input.nextInt(); D2 = input.nextInt();if(W1[0] == 0) b1 = solve(input, W1);if(W2[0] == 0) b2 = solve(input, W2);W[0] = W1[0] + W2[0];//System.out.println(W[0]);return b1 && b2 && (W1[0] * D1 == W2[0] * D2);}}

方法2：

import java.io.BufferedInputStream;import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner input = new Scanner(new BufferedInputStream(System.in));int T = input.nextInt();Weight w = new Weight();for(int i = 0; i < T; i++) {if(i != 0)System.out.println();if(solve(input, w))System.out.println("YES");elseSystem.out.println("NO");}}static class Weight {int w;public Weight(int w) {this. w = w;}public Weight() {super();}}public static boolean solve(Scanner input, Weight W) {Weight W1 = new Weight();Weight W2 = new Weight();int D1, D2;boolean b1 = true, b2 = true;W1.w = input.nextInt(); D1 = input.nextInt();W2.w = input.nextInt(); D2 = input.nextInt();if(W1.w == 0) b1 = solve(input, W1);if(W2.w == 0) b2 = solve(input, W2);W.w = W1.w + W2.w;//System.out.println(W.w);return b1 && b2 && (W1.w * D1 == W2.w * D2);}}