hdu4496 D-City

D-City

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

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
Hint
The graph given in sample input is a complete graph, that each pair of vertex has an edge connecting them, so there's only 1 connected block at first. The first 3 lines of output are 1s  because  after  deleting  the  first  3  edges  of  the  graph,  all  vertexes  still  connected together. But after deleting the first 4 edges of the graph, vertex 1 will be disconnected with other vertex, and it became an independent connected block. Continue deleting edges the disconnected blocks increased and finally it will became the number of vertex, so the last output should always be N.

反向并查集就可以了！

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define MAXN 10050
int father[MAXN],ans[MAXN];
struct edge{
    int s,e;
}l[100005];
int find(int x)
{
    if(father[x]!=x)
    father[x]=find(father[x]);
    return father[x];
}
int main()
{
   int n,i,m,all;
   while(scanf("%d%d",&n,&m)!=EOF)
   {
       all=n;
       for(i=0;i<=n;i++)
       {
           father[i]=i;
       }
       for(i=m-1;i>=0;i--)
       {
           scanf("%d%d",&l[i].s,&l[i].e);
       }
       for(i=0;i<m;i++)
       {
           int a=find(l[i].s);
           int b=find(l[i].e);
           if(a!=b)
           {
               father[a]=b;
               all--;
           }
           ans[i]=all;
       }
      // ans[1]=n;
       for(i=m-2;i>=0;i--)
       {
           printf("%d\n",ans[i]);
       }
       printf("%d\n",n);
   }
    return 0;
}