LA 3644 (并查集)

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：简单来说就是判断这些点有没有构成环，如果有就答案+1.

题解：使用并查集维护点信息，千万不要想着使用Floyd判断有没有环，这些点的个数都没有明确给出。这里我想的是使用按秩合并（按权值合并），其实不是用按秩合并也是可以的，只要每次都找到根，合并就可以了。自己想多了。。。。。。。

#include<iostream>#include<cstdio>#include<map>#include<algorithm>#include<string>#include<cmath>#include<vector>#include<queue>#include<deque>#include<cstring>#include<set>#include<vector>using namespace std;#define LL long long #define N 100005int fa[N];int Rank[N];inline void init(){for(int i=0;i<N;i++){fa[i]=i;Rank[i]=1;}}int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}int unio(int x,int y){x=find(x);y=find(y);if(x==y){return 1;}else{if(Rank[x]>Rank[y]){fa[y]=x;Rank[x]+=Rank[y];}else{fa[x]=y;Rank[y]+=Rank[x];}return 0;}}int main(){#ifdef CDZSCfreopen("i.txt","r",stdin);#endifinit();int ans=0;int a,b;while(~scanf("%d",&a)){if(a==-1){printf("%d\n",ans);init();ans=0;}else{scanf("%d",&b);ans+=unio(a,b);}}return 0;}