D-City

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)

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.

链接：http://acm.hdu.edu.cn/showproblem.php?pid=4496

题意：给你 n个点 m条边，首先边是都建好了的。 然后删一条边，然后问你当前图里有几个团。

做法：倒着做，起始5个点 ，5个团，每次先输出几个团。 然后开始并那两个给出的点，用并查集，然后find找到的祖先如果不同，那么就是有两个团并掉了。就把总的团数-1。 如果两个点 find 返回值相同 就是这两个点本来就是在一个团里的，就不用处理。每次先输出当前有几团再并。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>
#define INF 999999999
#define eps 0.00001
#define LL __int64d
#define pi acos(-1.0)

int n,m;
int lian[100010][2];
int f[10010];
int find(int x){return x==f[x]?x:f[x] = find(f[x]);}  //初始化为自己
int Union(int x, int y){
	int fx = find(x), fy = find(y);
	if(fx == fy)
		return 0;
	if(fx>fy)
		swap(fx,fy);
	f[fx] = f[x] = f[y] = fy;
	return 1;
}
int ans[100010];

int main()
{
	while(scanf("%d%d",&n,&m)!=EOF){
		for(int i=0;i<n;i++)
		{
			f[i]=i;
		}
		for(int i=0;i<m;i++)
		{
			scanf("%d%d",&lian[i][0],&lian[i][1]);
		}
		int nw=n;
		for(int i=m-1;i>=0;i--)
		{
			int tem=Union(lian[i][0],lian[i][1]);
			ans[i]=nw;
			if(tem)
				nw--;
		}
		for(int i=0;i<m;i++)
		 printf("%d\n",ans[i]);
	}
	return 0;
}