题目大意:维护一个森林,支持边的断,连,以及查询连通性
LCT裸题 洛谷P2147传送门
1A了,给自己鼓鼓掌
#include <cstdio>
#include <algorithm>
#include <cstring>
#define il inline
#define inf 500000
#define N 10100
using namespace std;
int n,m,tp;
int stk[N];
struct LinkCutTree{
int fa[N],ch[N][2],rv[N];
il int idf(int x){return ch[fa[x]][0]==x?0:1;}
il void con(int x,int ff,int p){fa[x]=ff,ch[ff][p]=x;}
il int isroot(int x){return (ch[fa[x]][0]!=x&&ch[fa[x]][1]!=x)?1:0;}
il void rev(int x){swap(ch[x][0],ch[x][1]),rv[x]^=1;}
il void pushdown(int x)
{
if(!rv[x]) return;
if(ch[x][0]) rev(ch[x][0]);
if(ch[x][1]) rev(ch[x][1]);
rv[x]=0;
}
il void rot(int x)
{
int y=fa[x];int ff=fa[y];int px=idf(x);int py=idf(y);
if(!isroot(y)) ch[ff][py]=x;
fa[x]=ff;con(ch[x][px^1],y,px),con(y,x,px^1);
//pushup(y),pushup(x);
}
void splay(int x)
{
int y=x;stk[++tp]=x;
while(!isroot(y)){stk[++tp]=fa[y],y=fa[y];}
while(tp){pushdown(stk[tp--]);}
while(!isroot(x))
{
y=fa[x];
if(isroot(y)) rot(x);
else if(idf(y)==idf(x)) rot(y),rot(x);
else rot(x),rot(x);
}
}
void access(int x){for(int y=0;x;y=x,x=fa[x])splay(x),ch[x][1]=y;/*pushup(x)*/}
il void mkroot(int x){access(x),splay(x),rev(x);}
il void split(int x,int y){mkroot(x),access(y),splay(y);}
int findrt(int x){access(x),splay(x);while(ch[x][0])pushdown(x),x=ch[x][0];return x;}
il void link(int x,int y){mkroot(x);if(findrt(y)!=x)fa[x]=y;}
il void cut(int x,int y){mkroot(x);if(findrt(y)==x&&fa[x]==y&&!ch[x][1])fa[x]=ch[y][0]=0;/*pushup(y)*/}
}lct;
int gint()
{
int rett=0,fh=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')fh=-1;c=getchar();}
while(c>='0'&&c<='9'){rett=(rett<<3)+(rett<<1)+c-'0';c=getchar();}
return rett*fh;
}
int main()
{
n=gint(),m=gint();
int fl,x,y;char str[20];
for(int i=1;i<=m;i++)
{
scanf("%s",str);
x=gint(),y=gint();
if(str[0]=='C'){lct.link(x,y);}
if(str[0]=='D'){lct.cut(x,y);}
if(str[0]=='Q'){
if(lct.findrt(x)==lct.findrt(y))
printf("Yes\n");
else printf("No\n");}
}
return 0;
}