洞穴勘测

luogu 2147

金牌导航 LCT-1

题目大意

给你若干操作，有三种操作：
1.连接两个点
2.吧两个点之间的连边断掉（保证有这条边）
3.查询两个点之间是否连通

样例 #1

输入样例 #1

200 5
Query 123 127
Connect 123 127
Query 123 127
Destroy 127 123
Query 123 127

输出样例 #1

No
Yes
No

样例#2

输入样例 #2

3 5
Connect 1 2
Connect 3 1
Query 2 3
Destroy 1 3
Query 2 3

输出样例 #2

Yes
No

数据范围

$1⩽n⩽10^4,m⩽2\times 10^5$

解题思路

在LCT上建出相应的图，每次查询判断根节点是否相同即可

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 10010
using namespace std;
int n, m, x, y, top, p[N], d[N], fa[N], son[N][2];
string str;
bool IR(int x)//LCT模板
{if (!fa[x]) return true;return son[fa[x]][0] != x && son[fa[x]][1] != x;
}
void down(int x)
{if (p[x] && x){swap(son[x][0], son[x][1]);p[son[x][0]] ^= 1;p[son[x][1]] ^= 1;p[x] = 0;}return;
}
bool ILS(int x)
{return x == son[fa[x]][0];
}
int which(int x)
{return x == son[fa[x]][1];
}
void rotate(int x)
{int y = fa[x], z = fa[y], k = !ILS(x), g = son[x][!k];if (!IR(y)) son[z][!ILS(y)] = x;son[x][!k] = y;son[y][k] = g;if (g) fa[g] = y;fa[x] = z;fa[y] = x;return;
}
void Splay(int x)
{d[top = 1] = x;for (int i = x; !IR(i); i = fa[i]) d[++top] = fa[i];while(top) down(d[top--]);while(!IR(x)){if (!IR(fa[x])){if (ILS(x) == ILS(fa[x])) rotate(fa[x]);else rotate(x);}rotate(x);}return;
}
void access(int x)
{for (int y = 0; x; y = x, x = fa[x]) Splay(x), son[x][1] = y;return;
}
void make_root(int x)
{access(x);Splay(x);p[x] ^= 1;return; 
}
int find_root(int x)
{access(x);Splay(x);down(x);while(son[x][0]) x = son[x][0], down(x);Splay(x);return x;
}
void link(int x, int y)
{make_root(x);fa[x] = y;return;
}
void cut(int x, int y)
{make_root(x);access(y);Splay(y);son[y][0] = 0;fa[x] = 0;return;
}
int main()
{scanf("%d%d", &n, &m);while(m--){cin>>str;scanf("%d%d", &x, &y);if (str == "Connect") link(x, y);else if (str == "Destroy") cut(x, y);else if (find_root(x) == find_root(y)) puts("Yes");//判断根节点是否相同else puts("No"); }return 0;
}