UVA 1160 - X-Plosives 即LA3644 并查集判断是否存在环

X-Plosives

A secret service developed a new kind ofexplosive that attain its volatile property only when a specific association ofproducts occurs. Each product is a mix of two different 易做图 compounds, towhich we call abinding pair. If N>2, thenmixing N different binding pairs containing N 易做图 compounds creates apowerful explosive.For example, the bindingpairs 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 guyin a delivery agency with one dangerous problem: receive binding pairs insequential order and place them in a cargo ship. However, you must avoidplacing in the same room an explosive association. So, after placing a set ofpairs, if you receive one pair that might produce an explosion with some of thepairs already in stock, you must refuse it, otherwise, you must accept it.

An example. Let’s assume you receive thefollowing sequence: A+B, G+B, D+F, A+E, E+G, F+H. You would accept the firstfour pairs but then refuse E+G since it would be possible to make the followingexplosive with the previous pairs: A+B, G+B, A+E, E+G (4 pairs with 4 易做图compounds). Finally, you would accept the last pair, F+H.

Compute thenumber 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 integersto 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 aline with the number –1. You may assume that no repeated binding pairsappears in the input.

Output

For each test case, a single line with the number ofrefusals.

Sample Input

1 2

3 4

3 5

3 1

2 3

4 1

2 6

6 5

-1

Sample Output

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（参考刘汝佳入门经典）

[cpp]
#include<stdio.h>
#include<string.h>
const int size=100001;
int parent[size],rank[size],count;
void init()
{
    int i;
    for(i=0;i<size;i++)
    {
        parent[i]=i;
        rank[i]=1;
    }
    count=0;
}
int find(int n)
{
    return n==parent[n]?n:parent[n]=find(parent[n]);
}
void join(int a,int b)
{
        if(rank[a]>rank[b])
        {
            parent[b]=a;
            rank[a]+=rank[b];
        }
        else
        {
            parent[a]=b;
            rank[b]+=rank[a];
        }
}
int main()
{
    int a,b;
    init();
    while(scanf("%d",&a)!=EOF)
    {
        if(a==-1)
        {
             printf("%d\n",count); init(); continue;
        }
        scanf("%d",&b);
        a=find(a);
        b=find(b);
        if(a==b) count++;
        else join(a,b);
    }
    return 0;
}

#include<stdio.h>
#include<string.h>
const int size=100001;
int parent[size],rank[size],count;
void init()
{
int i;
for(i=0;i<size;i++)
{
  parent[i]=i;
  rank[i]=1;
}
count=0;
}
int find(int n)
{
return n==parent[n]?n:parent[n]=find(parent[n]);
}
void join(int a,int b)
{
        if(rank[a]>rank[b])
  {
   parent[b]=a;
   rank[a]+=rank[b];
  }
  else
  {
   parent[a]=b;
   rank[b]+=rank[a];
  }
}
int main()
{
int a,b;
init();
while(scanf("%d",&a)!=EOF)
{
  if(a==-1)
  {
             printf("%d\n",count); init(); continue;
  }
  scanf("%d",&b);
        a=find(a);
  b=find(b);
  if(a==b) count++;
  else join(a,b);
}
return 0;
}

补充：软件开发 , C++ ,