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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. 给你一张图，问删除前i条边时的联通个数是多少？ 答案肯定是非递减序列，而且最后的答案肯定是n。我们知道，一开始的所有的点没有边时，也是n个连通块（每个点都是一个连通块）。连上最后一条边，连通块减一。这样我们倒着去找，思路就很清楚了，就是看连上这条边之后，连通块个数变为多少。 代码如下：

#pragma comment(linker, "/STACK:102400000,102400000")//这一块可以不用，一开始数组开小了#include
   
    #define ll long longusing namespace std;const int maxx=1e4+100;const int maxm=1e6+100;struct node{
   	int x,y;}p[maxm];int f[maxx],ans[maxm],size[maxx];int n,m,sum;inline int find(int u){
   	if(u==f[u]) return u;	else return f[u]=find(f[u]);}inline int merge(int u,int v){
   	int t1=find(u);	int t2=find(v);	if(t1!=t2)	{
   		sum--;		if(size[t1]
    
     =1;i--)		{
   			x=p[i].x,y=p[i].y;			if(merge(x,y)) ans[i]=sum+1;			else ans[i]=sum;		}		for(int i=1;i<=m;i++) printf("%d\n",ans[i]);	}	return 0;}
    
   

努力加油a啊，(^o)/~

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