Not so Mobile
Before being an ubiquous communications gadget, amobilewas 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 isWl×Dl=Wr×DrwhereDlis the left distance,Dris the right distance,Wlis the left weight andWris 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:WlDlWrDr
IfWlorWris 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 bothWlandWrare 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 Input10 2 0 40 3 0 11 1 1 12 4 4 21 6 3 2
Sample OutputYES
Jose Paulo Leal, ACM-UP’2001
#include<stdio.h>#include<malloc.h>#include<string.h>/*typedef struct BiTree{int weight, value;struct BiTree *lch, *rch;}BiTree, BiNTree;*/int Traverse(int& balance){int wl, dl, wr, dr;scanf(, &wl, &dl, &wr, &dr);/* BiNTree left, right;Tree->lch = (BiNTree)malloc(sizeof(BiTree));Tree->lch->weight = dl;Tree->lch->lch = Tree->lch->rch = NULL;if(wl == 0)Tree->lch->value = Traverse(Tree->lch, balance);else Tree->lch->value = wl;Tree->rch = (BiNTree)malloc(sizeof(BiTree));Tree->rch->weight = dr;Tree->rch->lch = Tree->rch->rch = NULL;if(wr == 0)Tree->rch->value = Traverse(Tree->rch, balance);else Tree->rch->value = wr;*/if(wl == 0) wl = Traverse(balance);if(wr == 0) wr = Traverse(balance);if(wl*dl != wr*dr) balance = 0;return wl+wr;}int main(){int T, n, i, j, balance;scanf(, &T);while(T–){/*Tree = (BiNTree)malloc(sizeof(BiTree));Tree->weight = Tree->value = NULL;Tree->lch = Tree->rch = NULL;*/balance = 1;Traverse(balance););););}return 0;}
解题思路：
题目的意思是：类似于杠杆原理你要保持平衡那么左边的重量乘以距离 == 右边的重量乘以右边的距离，只不过这里的意思是说如果mobile的左边或右边的重量变为零了，服务器空间，那么它下面坑定还有另外一辆mobile，香港空间，此时的重量替代为下面这辆mobile的左边和右边重量的相加，
赤裸裸的水题啊，害的我要看题目两边，而且代码中可以看到，被坑加忽悠了！！！
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