UVa 839 Not so Mobile(树的递归输入)
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题意 判断一个树状天平是否平衡 每个测试样例每行4个数 wl,dl,wr,dr 当wl*dl=wr*dr时 视为这个天平平衡 当wl或wr等于0是 下一行将是一个子天平 如果子天平平衡 wl为子天平的wl+wr 否则整个天平不平衡
容易看出 输入是递归的 所以我们可以直接递归 边输入边判断
#include<cstdio>using namespace std;bool solve(int &w){ int wl, dl, wr, dr; bool mobl = true, mobr = true; scanf("%d %d %d %d", &wl, &dl, &wr, &dr); if(!wl) mobl = solve(wl); if(!wr) mobr = solve(wr); w = wl + wr; return mobl && mobr && (wl * dl == wr * dr);}int main(){ int cas, w; scanf("%d", &cas); while(cas--) { printf(solve(w) ? "YES\n" : "NO\n"); if(cas) printf("\n"); } return 0;}
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×Drwhere 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 Wrare 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
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×Drwhere 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 Wrare 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
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