UVALive

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 10 5 ) 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：有若干个由两种元素组成的简单化合物，现在把它们装进车里，如果车里有k种简单化合物并且在这k种简单化合物中恰好有k种元素的话，那么就会引发爆炸，所以车上的化合物必须避免满足这个条件。求出这些化合物中有多少个化合物不能装进车。

分析：最开始没有找到规律，直到后来用图的思想去思考后才发现规律，发生爆炸的条件就是形成环。怎么判断有没有形成环呢？可以用并查集构造树，如果即将合并的两个结点有相同的根，则会形成环。

AC代码如下（3ms） ：

#include <stdio.h>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
#define fast                                ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ll                                  long long

const int maxn = 1e5 + 5;
int par[maxn];

void init(int n)
{
	for (int i = 0; i < maxn; i++)
		par[i] = i;
}

int find_(int x)
{
	return (par[x] == x) ? x : par[x] = find_(par[x]);
}

int main()
{
	int x,y;
	while (~scanf("%d", &x))
	{
		int ans = 0;
		init(maxn);
		while (x != -1)
		{
			scanf("%d", &y);
			int rx = find_(x);
			int ry = find_(y);
			if (rx == ry)ans++;//如果即将连接的两个结点拥有相同的根，则会成环
			else par[rx] = ry;
			scanf("%d", &x);
		}
		printf("%d
", ans);
	}
	return 0;
}