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

 

Sample Output

 

Source

 

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liuyiding

 

 /* ***********************************************

 Author        :kuangbin

 Created Time  :2013/8/24 12:20:44

 File Name     :F:\2013ACM练习\比赛练习\2013通化邀请赛\1004.cpp

 ************************************************ */

 #include <stdio.h>

 #include <string.h>

 #include <iostream>

 #include <algorithm>

 #include <vector>

 #include <queue>

 #include <set>

 #include <map>

 #include <string>

 #include <math.h>

 #include <stdlib.h>

 #include <time.h>

 using namespace std;

 const int MAXN = ;

 int F[MAXN];

 int find(int x)

 {

     if(F[x] == -)return x;

     return F[x] = find(F[x]);

 }

 pair<int,int>p[];

 int ans[];

 int main()

 {

     //freopen("in.txt","r",stdin);

     //freopen("out.txt","w",stdout);

     int n,m;

     while(scanf("%d%d",&n,&m) == )

     {

         memset(F,-,sizeof(F));

         for(int i = ;i <= m;i++)

         {

             scanf("%d%d",&p[i].first,&p[i].second);

         }

         ans[m] = n;

         for(int i = m;i > ;i --)

         {

             int t1 = find(p[i].first);

             int t2 = find(p[i].second);

             if(t1 != t2)

             {

                 ans[i-] = ans[i] - ;

                 F[t1] = t2;

             }

             else ans[i-] = ans[i];

         }

         for(int i = ;i <= m;i++)

             printf("%d\n",ans[i]);

     }

     return ;

 }