LCT判断两点联通的这样的一个基础问题,因为不存在环,所以直接LCT维护连接关系即可。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e4 + 7;
int N, M;
namespace LCT
{
int fa[maxN], c[maxN][2];
int r[maxN];
bool isroot(int x) { return c[fa[x]][0] != x && c[fa[x]][1] != x; }
void pushup(int x)
{
}
void pushr(int x) { swap(c[x][0], c[x][1]); r[x] ^= 1; }
void pushdown(int x)
{
if(r[x])
{
if(c[x][0]) pushr(c[x][0]);
if(c[x][1]) pushr(c[x][1]);
r[x] = 0;
}
}
void Rotate(int x)
{
int y = fa[x], z = fa[y], k = c[y][1] == x;
if(!isroot(y)) c[z][c[z][1] == y] = x;
fa[x] = z;
c[y][k] = c[x][k ^ 1];
fa[c[x][k ^ 1]] = y;
c[x][k ^ 1] = y;
fa[y] = x;
pushup(y);
pushup(x);
if(z) pushup(z);
}
int Stap[maxN];
void Splay(int x)
{
int y = x, z = 0;
Stap[++z] = y;
while(!isroot(y)) Stap[++z] = y = fa[y];
while(z) pushdown(Stap[z--]);
while(!isroot(x))
{
y = fa[x]; z = fa[y];
if(!isroot(y)) (c[z][0] == y) ^ (c[y][0] == x) ? Rotate(x) : Rotate(y);
Rotate(x);
}
}
void Access(int x)
{
int y = 0;
while(x)
{
Splay(x);
c[x][1] = y;
pushup(x);
y = x;
x = fa[x];
}
}
void makeroot(int x)
{
Access(x);
Splay(x);
pushr(x);
}
int findroot(int x)
{
Access(x);
Splay(x);
while(c[x][0])
{
pushdown(x);
x = c[x][0];
}
Splay(x);
return x;
}
void Split(int x, int y)
{
makeroot(x);
Access(y);
Splay(y);
}
void link(int x, int y)
{
makeroot(x);
if(findroot(y) != x)
{
fa[x] = y;
}
}
void cut(int x, int y)
{
makeroot(x);
if(findroot(y) != x || fa[y] != x || c[y][0]) return;
fa[y] = c[x][1] = 0;
pushup(x);
}
};
using namespace LCT;
int main()
{
scanf("%d%d", &N, &M);
char ch[10]; int u, v;
while(M--)
{
scanf("%s%d%d", ch, &u, &v);
switch (ch[0])
{
case 'Q':
{
if(findroot(u) == findroot(v)) printf("Yes\n");
else printf("No\n");
break;
}
case 'C':
{
link(u, v);
break;
}
default:
{
cut(u, v);
break;
}
}
}
return 0;
}