X-Plosives
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. Let’s 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. The input
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, 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
并查集！！
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100000 + 10;
int pa[maxn];
int find(int x)
{ //并查集的查找操作，带路径压缩
return pa[x] != x ? pa[x] = find(pa[x]) : x;
}
int main()
{
int x, y;
while(scanf("%d", &x) != EOF)
{
for(int i = 0; i < maxn; i++) pa[i] = i; //并查集首先要初始化
int ref = 0;
while(x != -1)
{
scanf("%d", &y);
x = find(x); y = find(y); //这之后，x，y分别是两个集合的代表元
if(x == y) ref++; //如果x和y代表同一个集合，则拒绝
else pa[x] = y; //否则合并，注意这里不是启发式合并
scanf("%d", &x);
}
printf("%d\n", ref);
}
return 0;
}