uva 839 not so mobile——yhx

Posted 2020-06-20 报告振兴哥

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.

 

 

 

(picture copy failed,cou huo zhe kan ba.)

 

 

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.

 1 #include<cstdio>
 2 bool slv(int &x)   //读入和处理同时进行
 3 {                  //变量不是从上往下传，而是从下往上传。
 4     int i,j,k,wl,dl,wr,dr;
 5     bool b1=1,b2=1;
 6     scanf("%d%d%d%d",&wl,&dl,&wr,&dr);
 7     if (!wl) b1=slv(wl);   //判定子问题的同时求出w1
 8     if (!wr) b2=slv(wr);
 9     x=wl+wr;         //对于本层递归没有意义，但为上一层传值。
10     if (b1&&b2&&wl*dl==wr*dr) return 1;
11     else return 0;
12 }
13 int main()
14 {
15     int i,n,x;
16     scanf("%d",&n);
17     for (i=1;i<=n;i++)
18     {
19         if (slv(x)) printf("YES\n");
20         else printf("NO\n");
21         if (i!=n) printf("\n");
22     }
23 }

极其精简的代码。算法没什么，具体实现见注释。