City -hdoj 4496-并查集删边

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.

InputFirst 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. 
OutputOutput 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

算法：

//正常并查集是一条一条边的加，然后判断联通块，但这个刚好反过来，
//那么直接储存反过来记录查询就行了
#include   
#include   
#define INF 0x3f3f3f3f 
#define max1 100000+50
#define max2 10000+50
int par[max2],a[max1];
using namespace std;
struct Node{
	int a,b;	
}data[max1];
int find(int x)
{
	if(x!=par[x])
		return par[x]=find(par[x]);
	return x;	
}
void join(int a, int b)
{
	int fa = find(a);
	int fb = find(b);
	if (fa != fb)
		par[fa] = fb;
}

int main()
{
	int m,n;
	int ans;
	while(scanf("%d %d",&n,&m)!=EOF)
	{
		ans=n;//最初n个城市，连通块个数为n 
		for(int i = 0; i < m; i++)		
		scanf("%d %d",&data[i].a,&data[i].b);
		for(int i = 0; i < n; i++)
		par[i]=i;
		for(int i=m-1; i>=0;i--)//倒序查询合并 
	        {
	            a[i]=ans;
	            if(find(data[i].a)!=find(data[i].b))
	            {
	                ans--;//如果不连通就把他们连起来，个数减一 
	            }               
	            join(data[i].a,data[i].b);
	        }
	        for (int i=0;i