【BZOJ2049】[SDOI2008] Cave 洞穴勘测(LCT维护连通性)
大致题意: 有\(n\)个洞穴,\(3\)种操作:连一条边,删一条边,询问两点是否联通。
\(LCT\)维护连通性
这道题应该是\(LCT\)动态维护连通性的一道模板题。
考虑将\(x\)和\(y\)连边时,我们就在\(LCT\)上\(Link(x,y)\)。
同理,\(x\)和\(y\)断边时,就\(Cut(x,y)\)。
询问连通性时,只要判断\(FindRoot(x)\)与\(FindRoot(y)\)是否相等即可。
代码
#include<bits/stdc++.h>
#define N 10000
#define swap(x,y) (x^=y^=x^=y)
using namespace std;
int n;
class Class_FIO
{
private:
#define Fsize 100000
#define tc() (A==B&&(B=(A=Fin)+fread(Fin,1,Fsize,stdin),A==B)?EOF:*A++)
char ch,*A,*B,Fin[Fsize];
public:
Class_FIO() {A=B=Fin;}
inline void read(int &x) {x=0;while(!isdigit(ch=tc()));while(x=(x<<3)+(x<<1)+(ch&15),isdigit(ch=tc()));}
inline void reads(string &x) {x="";while(isspace(ch=tc()));while(x+=ch,!isspace(ch=tc()));}
}F;
class Class_LCT//LCT模板
{
private:
#define LCT_SIZE N
#define Rever(x) (swap(node[x].Son[0],node[x].Son[1]),node[x].Rev^=1)
#define PushDown(x) (node[x].Rev&&(Rever(node[x].Son[0]),Rever(node[x].Son[1]),node[x].Rev=0))
#define Which(x) (node[node[x].Father].Son[1]==x)
#define Connect(x,y,d) (node[node[x].Father=y].Son[d]=x)
#define IsRoot(x) (node[node[x].Father].Son[0]^x&&node[node[x].Father].Son[1]^x)
#define MakeRoot(x) (Access(x),Splay(x),Rever(x))
int Stack[LCT_SIZE+5];
struct Tree
{
int Father,Rev,Son[2];
}node[LCT_SIZE+5];
inline void Rotate(int x)
{
register int fa=node[x].Father,pa=node[fa].Father,d=Which(x);
!IsRoot(fa)&&(node[pa].Son[Which(fa)]=x),node[x].Father=pa,Connect(node[x].Son[d^1],fa,d),Connect(fa,x,d^1);
}
inline void Splay(int x)
{
register int fa=x,Top=0;
while(Stack[++Top]=fa,!IsRoot(fa)) fa=node[fa].Father;
while(Top) PushDown(Stack[Top]),--Top;
while(!IsRoot(x)) fa=node[x].Father,!IsRoot(fa)&&(Rotate(Which(x)^Which(fa)?x:fa),0),Rotate(x);
}
inline void Access(int x) {for(register int son=0;x;x=node[son=x].Father) Splay(x),node[x].Son[1]=son;}
inline int FindRoot(int x) {Access(x),Splay(x);while(node[x].Son[0]) PushDown(x),x=node[x].Son[0];return Splay(x),x;}
public:
inline void Link(int x,int y) {MakeRoot(x),FindRoot(y)^x&&(node[x].Father=y);}
inline void Cut(int x,int y) {MakeRoot(x),!(FindRoot(y)^x)&&!(node[y].Father^x)&&!node[y].Son[0]&&(node[y].Father=node[x].Son[1]=0);}
inline bool IsConnected(int x,int y) {return !(FindRoot(x)^FindRoot(y));}//判断连通性
}LCT;
int main()
{
register int query_tot,x,y;register string op;
for(F.read(n),F.read(query_tot);query_tot;--query_tot)
{
F.reads(op),F.read(x),F.read(y);
switch(op[0])
{
case 'C':LCT.Link(x,y);break;//连边
case 'D':LCT.Cut(x,y);break;//删边
case 'Q':puts(LCT.IsConnected(x,y)?"Yes":"No");break;//询问
}
}
return 0;
}
待到再迷茫时回头望,所有脚印会发出光芒