Codeforces 292D. Connected Components

Codeforces 292D. Connected Components

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D. Connected Components time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

We already know of the large corporation where Polycarpus works as a system administrator. The computer network there consists of ncomputers and m cables that connect some pairs of computers. In other words, the computer network can be represented as some non-directed graph with n nodes and m edges. Let's index the computers with integers from 1 to n, let's index the cables with integers from 1 tom.

Polycarpus was given an important task — check the reliability of his company's network. For that Polycarpus decided to carry out a series of k experiments on the computer network, where the i-th experiment goes as follows:

1. Temporarily disconnect the cables with indexes from li to ri, inclusive (the other cables remain connected).
2. Count the number of connected components in the graph that is defining the computer network at that moment.
3. Re-connect the disconnected cables with indexes from li to ri (that is, restore the initial network).

   Help Polycarpus carry out all experiments and for each print the number of connected components in the graph that defines the computer network through the given experiment. Isolated vertex should be counted as single component.

   Input

   The first line contains two space-separated integers n, m (2?≤?n?≤?500; 1?≤?m?≤?104) — the number of computers and the number of cables, correspondingly.

   The following m lines contain the cables' description. The i-th line contains space-separated pair of integers xi, yi (1?≤?xi,?yi?≤?n; xi?≠?yi)— the numbers of the computers that are connected by the i-th cable. Note that a pair of computers can be connected by multiple cables.

   The next line contains integer k (1?≤?k?≤?2·104) — the number of experiments. Next k lines contain the experiments' descriptions. The i-th line contains space-separated integers li, ri (1?≤?li?≤?ri?≤?m) — the numbers of the cables that Polycarpus disconnects during the i-th experiment.

   Output

   Print k numbers, the i-th number represents the number of connected components of the graph that defines the computer network during the i-th experiment.

   Sample test(s) input

   6 5
   1 2
   5 4
   2 3
   3 1
   3 6
   6
   1 3
   2 5
   1 5
   5 5
   2 4
   3 3
   

   output

   4
   5
   6
   3
   4
   2

   
   

   
   

   
   

   /**
    * Created by ckboss on 14-9-8.
    */
   import java.util.*;
   import java.io.*;
   
   public class ConnectedComponents {
   
       static class Edge{
           int u,v;
       }
       static final int maxn=11000;
   
       static Edge[] edge=new Edge[maxn];
       static int[] fa=new int[600];
       static int n,m,q;
   
       static int Find(int x){
           if(x==fa[x]){
               return x;
           }
           return fa[x]=Find(fa[x]);
       }
   
       static boolean Union(int x,int y){
           int X=Find(x),Y=Find(y);
           if(X==Y) return false;
           fa[X]=Y;
           return true;
       }
   
       static int[] front=new int[maxn],back=new int[maxn];
       static int nf,nb;
   
   
       public static void main(String[] args){
   
           InputStream inputStream = System.in;
           OutputStream outputStream = System.out;
           Scanner in = new Scanner(inputStream);
           PrintWriter out = new PrintWriter(outputStream);
   
           n=in.nextInt(); m=in.nextInt();
   
           for(int i=1;i<=m;i++){
               edge[i]=new Edge();
               edge[i].u=in.nextInt();
               edge[i].v=in.nextInt();
           }
   
           for(int i=1;i<=n;i++) fa[i]=i;
           for(int i=1;i<=m;i++){
               if(Union(edge[i].u,edge[i].v)==true){
                   front[nf++]=i;
               }
           }
   
           for(int i=1;i<=n;i++) fa[i]=i;
           for(int i=m;i>=1;i--){
               if(Union(edge[i].u,edge[i].v)==true){
                   back[nb++]=i;
               }
           }
   
           q=in.nextInt();
   
           while(q-->0){
               int l,r;
               l=in.nextInt();r=in.nextInt();
   
               for(int i=1;i<=n;i++){
                   fa[i]=i;
               }
   
               int t=n;
               for(int i=0;ir;i++){
                   if(Union(edge[back[i]].u,edge[back[i]].v)==true){
                       t--;
                   }
               }
   
               System.out.println(t);
           }
   
       }
   }