HDU4496(并查集)

D-City

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3665    Accepted Submission(s): 1306

Problem Description

Luxer is a really bad guy. He destroys everything he met. 
One day Luxer went to D-city. D-city has N D-points and M D-lines. Each D-line connects exactly two D-points. Luxer will destroy all the D-lines. The mayor of D-city wants to know how many connected blocks of D-city left after Luxer destroying the first K D-lines in the input. 
Two points are in the same connected blocks if and only if they connect to each other directly or indirectly.

 

Input

First line of the input contains two integers N and M. 
Then following M lines each containing 2 space-separated integers u and v, which denotes an D-line. 
Constraints: 
0 < N <= 10000 
0 < M <= 100000 
0 <= u, v < N. 

 

Output

Output M lines, the ith line is the answer after deleting the first i edges in the input.

 

Sample Input

5 10

0 1

1 2

1 3

1 4

0 2

2 3

0 4

0 3

3 4

2 4

 

Sample Output

1

1

1

2

2

2

2

3

4

5

 

 并查集，题目问每破坏一条边有多上连通块，可以倒推把边连起来，每次查询有多少个集合

 //2016.8.9
 #include<iostream>
 #include<cstdio>
 #include<stack>

 using namespace std;

 int fa[], arr[][], cnt;

 int getf(int i)
 {
     if(fa[i]==i)return i;
     return fa[i] = getf(fa[i]);
 }

 int init(int n)
 {
     for(int i = ; i < n; i++)
       fa[i] = i;
 }

 int Merge(int a, int b)
 {
     int af = getf(a);
     int bf = getf(b);
     if(af!=bf){
       fa[bf] = af;
       cnt--;
     }
 }

 int main()
 {
     int n, m, u, v;
     while(cin>>n>>m)
     {
         init(n);
         for(int i = ; i < m; i++)
         {
             scanf("%d%d", &arr[i][], &arr[i][]);
         }
         stack<int> s;
         cnt = n;
         for(int i = m-; i >= ; i--)
         {
             s.push(cnt);
             Merge(arr[i][], arr[i][]);
         }
         while(!s.empty())
         {
             int ans = s.top();
             cout<<ans<<endl;
             s.pop();
         }
     }

     return ;
 }