A secret service developed a new kind of explosive that attain its volatile property only when a specific 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 first 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n种化合物，每种化合物由两种元素组成。当几种的化合物数量等于他们所含不同元素的数量时，就会发生爆炸。现在依次给出化合物的组成，当新的化合物与之前的化合物放在一起会发生爆炸时，就不能允许这个化合物放进来。输出拒绝的次数。
思路：
把元素看成点，化合物看成边，每次新的化合物进来当成连一条边。如果图中没有环，则每个连通分量是一棵树，其边数等于点数减1，不可能存在爆炸的情况；如果图中有环，则这个环上点数等于边数，就会爆炸。使用并查集连边，如果要连的两个点在同一集合中，则答案加1。
code：
#include <iostream> #include<algorithm> #include <cstdio> #include<cstring> #include<math.h> #include<memory> using namespace std; typedef long long LL; #define max_v 100005 int pa[max_v]; int re[max_v]; int n,ans; void make_set(int x) { pa[x]=x; re[x]=0; } int find_set(int x) { if(x!=pa[x]) pa[x]=find_set(pa[x]); return pa[x]; } void union_set(int x,int y) { x=find_set(x); y=find_set(y); if(x==y) { ans++; return ; } if(re[x]>re[y]) pa[y]=x; else { pa[x]=y; if(re[x]==re[y]) re[y]++; } } int main() { int x,y; while(~scanf("%d",&x)) { ans=0; for(int i=1;i<max_v;i++) make_set(i); while(x!=-1) { scanf("%d",&y); union_set(x,y); scanf("%d",&x); } printf("%d\n",ans); } return 0; } //判断有没有构成环，构成环的不放进去且计数加1 //输出计数