【題解】

　　動態樹模板題，只要求維護森林的連通性，直接上板子即可。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #define N 500010
 4 #define ls (c[u][0])
 5 #define rs (c[u][1])
 6 using namespace std;
 7 int n,m;
 8 inline int read(){
 9     int k=0,f=1; char c=getchar();
10     while(c<‘0‘||c>‘9‘)c==‘-‘&&(f=-1
 
),c=getchar();
11     while(‘0‘<=c&&c<=‘9‘)k=k*10+c-‘0‘,c=getchar();
12     return k*f;
13 }
14 struct Link_Cut_Tree{
15     int top,c[N][2],fa[N],rev[N],q[N];
16     inline void pushdown(int u){if(rev[u]) rev[ls]^=1,rev[rs]^=1,rev[u]^=1,swap(ls,rs);}
17     inline bool isroot(int u){return
 
 c[fa[u]][0]!=u&&c[fa[u]][1]!=u;}
18     inline bool which(int u){return c[fa[u]][1]==u;}
19     void rotate(int u){
20         int f=fa[u],gf=fa[f],wh=which(u);
21         if(!isroot(f)){c[gf][which(f)]=u;}
22         fa[u]=gf; fa[f]=u; fa[c[u][wh^1]]=f;
23         c[f][wh]=c[u][wh^1]; c[u][wh^1
 
]=f;
24     }
25     void splay(int u){
26         top=1; q[top]=u;
27         for(int i=u;!isroot(i);i=fa[i]) q[++top]=fa[i];
28         for(int i=top;i;i--) pushdown(q[i]);
29         while(!isroot(u)){
30             if(!isroot(fa[u])) rotate(which(u)==which(fa[u])?fa[u]:u);
31             rotate(u);
32         }
33     }
34     void access(int u){for(int son=0;u;son=u,u=fa[u]) splay(u),c[u][1]=son;}
35     void makeroot(int u){
36         access(u); splay(u); rev[u]^=1;
37     }
38     int find(int u){
39         access(u); splay(u);
40         while(ls) u=ls; return u;
41     }
42     void split(int x,int y){
43         makeroot(x); access(y); splay(y); 
44     }
45     void cut(int x,int y){
46         split(x,y);
47         if(c[y][0]==x) c[y][0]=0,fa[x]=0;
48     }
49     void link(int x,int y){
50         makeroot(x); fa[x]=y;
51     }
52 }t;
53 int main(){
54     n=read(); m=read();
55     while(m--){
56         char c=getchar();
57         while(c!=‘C‘&&c!=‘D‘&&c!=‘Q‘) c=getchar();
58         char c2=getchar();
59         while(c2!=‘ ‘) c2=getchar();
60         int x=read(),y=read(),xrt=t.find(x),yrt=t.find(y);
61         if(c==‘C‘) if(xrt!=yrt) t.link(x,y);
62         if(c==‘D‘) if(xrt==yrt) t.cut(x,y);
63         if(c==‘Q‘) puts(xrt==yrt?"Yes":"No");
64     }
65     return 0;
66 }

洛谷 2147 SDOI2008 Cave 洞穴勘測