【题意分析】

　　给你一张无向图，要求支持删点和询问连通块数。

【解题思路】

　　可以直接可持久化并查集大力艹过去。

　　考虑到正着删点就是倒着加点，所以并不需要可持久化。复杂度O((k+m)α(n))。

【参考代码】

　　当时还在玩泥巴，实现有点naive，常数奇大无比。。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<set>
 4 #define REP(I,start,end) for(int I=start;I<=end;I++)
 5 #define PER(I,start,end) for(int I=start;I>=end;I--)
 6 using namespace std;
 7 set<int> tree;
 8 int n,m,k,edge=0,q[400001],father[400001],head[400001],next[400001],point[400001],sav[400001];
 9 bool destroyed[400001];
10 inline void addEdge(int from,int to)
11 {
12     next[++edge]=head[from];
13     head[from]=edge;
14     point[edge]=to;
15 }
16 inline int getFa(int n)
17 {
18     if(father[n]==n)
19         return n;
20     return father[n]=getFa(father[n]);
21 }
22 int main()
23 {
24     scanf("%d%d",&n,&m);
25     REP(i,1,m)
26     {
27         int u,v;
28         scanf("%d%d",&u,&v);
29         addEdge(u,v);
30         addEdge(v,u);
31     }
32     scanf("%d",&k);
33     memset(destroyed,0,sizeof(destroyed));
34     REP(i,1,k)
35     {
36         scanf("%d",q+i);
37         destroyed[q[i]]=true;
38     }
39     REP(i,0,n-1)
40         father[i]=i;
41     REP(i,0,n-1)
42         if(!destroyed[i])
43         {
44             int t=head[i];
45             while(t)
46             {
47                 int p=point[t];
48                 if(!destroyed[p])
49                     father[getFa(p)]=getFa(i);
50                 t=next[t];
51             }
52         }
53     tree.clear();
54     REP(i,0,n-1)
55     {
56         int fa=getFa(i);
57         if(!destroyed[i]&&tree.find(fa)==tree.end())
58             tree.insert(fa);
59     }
60     sav[k]=tree.size();
61     PER(i,k-1,0)
62     {
63         int P=q[i+1],t=head[P];
64         destroyed[P]=false;
65         tree.clear();
66         while(t)
67         {
68             int p=point[t];
69             if(!destroyed[p])
70             {
71                 int fa2=getFa(p);
72                 tree.insert(fa2);
73                 father[getFa(P)]=fa2;
74             }
75             t=next[t];
76         }
77         sav[i]=sav[i+1]-tree.size()+1;
78     }
79     REP(i,0,k)
80         printf("%d\n",sav[i]);
81     return 0;
82 }
View Code