A secret service developed a new kind of explosive that attain its volatile property only when a specic
association of products occurs. Each product is a mix of two different simple compounds, to which we
call a binding pair. If N > 2, then mixing N different binding pairs containing N simple compounds
creates a powerful explosive. For example, the binding pairs A+B, B+C, A+C (three pairs, three
compounds) result in an explosive, while A+B, B+C, A+D (three pairs, four compounds) does not.
You are not a secret agent but only a guy in a delivery agency with one dangerous problem: receive
binding pairs in sequential order and place them in a cargo ship. However, you must avoid placing in
the same room an explosive association. So, after placing a set of pairs, if you receive one pair that
might produce an explosion with some of the pairs already in stock, you must refuse it, otherwise, you
must accept it.
An example. Lets assume you receive the following sequence: A+B, G+B, D+F, A+E, E+G,
F+H. You would accept the rst four pairs but then refuse E+G since it would be possible to make the
following explosive with the previous pairs: A+B, G+B, A+E, E+G (4 pairs with 4 simple compounds).
Finally, you would accept the last pair, F+H.
Compute the number of refusals given a sequence of binding pairs.
Input
The input will contain several test cases, each of them as described below. Consecutive
test cases are separated by a single blank line.
Instead of letters we will use integers to represent compounds. The input contains several lines.
Each line (except the last) consists of two integers (each integer lies between 0 and 105
) separated by
a single space, representing a binding pair.
Each test case ends in a line with the number `-1'. You may assume that no repeated binding pairs
appears in the input.
Output
For each test case, the output must follow the description below.
A single line with the number of refusals.
Sample Input
1 2
3 4
3 5
3 1
2 3
4 1
2 6
6 5
-1
Sample Output
3

题意：   有一些简单化合物   每个化合物由2中不同元素组成        然后按照顺序依次将这些化合物放进车里   但是如果车上存在k个简单化合物 且正好包含k中元素的话

那么他们将变成易爆的化合物   为安全起见     每当你拿到一个化合物的时候   如果它和已装车的化合物形成易爆化合物   你就应当拒绝装车  否则就应该装车

请输出有多少个化合物没有装车

思路：

注意题目要求k个简单化合物 且正好包含k中元素   是任意k个化合物 也就是说 只要车里存在任意k个化合物 如果他们含有元素也为k个 则不能装入这样的一个化合物

可以把每个元素看成顶点   一个化合物作为一条边   当整个图存在环的时候 组成环的边对应的化合物就是危险的 否则是安全的

判断是否会组成环 可以通过并查集  如果要添加的边 x y同时在同一个集合中  那么它将组成环  拒绝它  

代码：

 #include<stdio.h>

 #include<string.h>

 #include<math.h>

 #include<ctype.h>

 #include<stdlib.h>

 #include<stdbool.h>

 #define rep(i,a,b)      for(i=(a);i<=(b);i++)

 #define clr(x,y)        memset(x,y,sizeof(x))

 #define sqr(x)          (x*x)

 int i,j,n,f1,f2,

     father[];

 void pre()

 {

     clr(father,);

     return ;

 }

 int find(int x)

 {

     int r,k,p;

     r=x;

     while(father[r]!=r) r=father[r];

     k=x;

     while(k!=father[k]) {

         p=father[k];

         father[k]=r;

         k=p;

     }

     return r;

 }

 int main()

 {

     int i,x,y,sum;

     pre();

     while(scanf("%d",&x)==){

         rep(i,,) father[i]=i;

         sum=;

         while(x!=-) {

             scanf("%d",&y);

             f1=find(x);f2=find(y);

             if(f1==f2) sum++;

             else father[f1]=f2;

             scanf("%d",&x);

         }

         printf("%d\n",sum);

     }

     return ;

 }