【BZOJ2049】[SDOI2008] Cave 洞穴勘测(LCT维护连通性)

点此看题面

大致题意:\(n\)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\(x\)\(y\)连边时,我们就在\(LCT\)\(Link(x,y)\)

同理,\(x\)\(y\)断边时,就\(Cut(x,y)\)

询问连通性时,只要判断\(FindRoot(x)\)\(FindRoot(y)\)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; 
}
posted @ 2018-12-19 19:50  TheLostWeak  阅读(...)  评论(... 编辑 收藏