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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 W_l×D_l = W_r×D_rwhere 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 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_rare zero then the following lines define two sub-mobiles: first the left then the right one.

Output 

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

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 W_l×D_l = W_r×D_rwhere 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_rare 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

UVa 839 Not so Mobile(树的递归输入)

标签：acm   uva   小白   数据结构   

原文地址：http://blog.csdn.net/acvay/article/details/39518985