Codeforces Round #266 (Div. 2) Problem E Solution

题目大意：

自己去看。。。

Solution:

直接离线进行处理，首先构建出完整的树，处理出dfs序。对于每个询问，事实上是问一个点是否在一条路径上，究竟是哪条路径可以利用并查集维护。

那么我们就需要知道一个点u是否是另一个点v的祖先。

事实上，这等价于：in[u]<=in[v]&&out[u]>=out[v].

于是总的时间复杂度为O(n\alpha(n)).

Code:

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cctype>
using namespace std;

inline int getc() {
	static const int L = 1 << 15;
	static char buf[L], *S = buf, *T = buf;
	if (S == T) {
		T = (S = buf) + fread(buf, 1, L, stdin);
		if (S == T)
			return EOF;
	}
	return *S++;
}
inline int getint() {
	int c;
	while(!isdigit(c = getc()));
	int tmp = c - '0';
	while(isdigit(c = getc()))
		tmp = (tmp << 1) + (tmp << 3) + c - '0';
	return tmp;
}

#define N 100010
int head[N], next[N], end[N];
void addedge(int a, int b) {
	static int q = 1;
	end[q] = b;
	next[q] = head[a];
	head[a] = q++;
}

int root[N], size[N];
int find(int x) {
	return x == root[x] ? x : root[x] = find(root[x]);
}

int down[N], up[N];

struct Ask {
	int ins, lab;
	Ask(int _ins = 0, int _lab = 0):ins(_ins),lab(_lab){}
}S[N];
int num;

int in[N], out[N], tclock;
bool vis[N];
void dfs(int x) {
	vis[x] = 1;
	in[x] = ++tclock;
	for(int j = head[x]; j; j = next[j])
		dfs(end[j]);
	out[x] = tclock;
}

inline bool inanc(int Son, int Anc) {
	return in[Anc] <= in[Son] && out[Anc] >= out[Son];
}

int main() {
	int n, m;
	n = getint();
	m = getint();
	register int i;
	
	for(i = 1; i <= n; ++i)
		root[i] = i, size[i] = 1;
		
	int sign, now = 0, a, b, x;
	for(i = 1; i <= m; ++i) {
		sign = getint();
		if (sign == 1) {
			a = getint(), b = getint();
			addedge(b, a);
			vis[a] = 1;
			root[a] = b;
		}
		if (sign == 2) {
			a = getint();
			down[++now] = a;
			up[now] = find(a);
		}
		if (sign == 3) {
			a = getint(), x = getint();
			S[++num] = Ask(a, x);
		}
	}
	
	for(i = 1; i <= n; ++i)
		if (!vis[i])
			dfs(i);
	
	for(i = 1; i <= num; ++i) {
		if (inanc(down[S[i].lab], S[i].ins) && inanc(S[i].ins, up[S[i].lab]))
			puts("YES");
		else
			puts("NO");
	}
	
	return 0;
}