Information Graph（并查集）

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本文介绍了一个使用并查集解决公司内部文件传递及查询问题的方法。在N个员工和M个事件中，通过维护两个并查集来跟踪文件的传递路径，确保能高效回答特定员工是否签署过某文件的问题。

There are n employees working in company "X" (let's number them from 1 to n for convenience). Initially the employees didn't have any relationships among each other. On each of m next days one of the following events took place:

either employee y became the boss of employee x (at that, employee x didn't have a boss before);
or employee x gets a packet of documents and signs them; then he gives the packet to his boss. The boss signs the documents and gives them to his boss and so on (the last person to sign the documents sends them to the archive);
or comes a request of type "determine whether employee x signs certain documents".

Your task is to write a program that will, given the events, answer the queries of the described type. At that, it is guaranteed that throughout the whole working time the company didn't have cyclic dependencies.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of employees and the number of events.

Each of the next m lines contains the description of one event (the events are given in the chronological order). The first number of the line determines the type of event t (1 ≤ t ≤ 3).

If t = 1, then next follow two integers x and y (1 ≤ x, y ≤ n) — numbers of the company employees. It is guaranteed that employee x doesn't have the boss currently.
If t = 2, then next follow integer x (1 ≤ x ≤ n) — the number of the employee who got a document packet.
If t = 3, then next follow two integers x and i (1 ≤ x ≤ n; 1 ≤ i ≤ [number of packets that have already been given]) — the employee and the number of the document packet for which you need to find out information. The document packets are numbered started from 1 in the chronological order.

It is guaranteed that the input has at least one query of the third type.

Output

For each query of the third type print "YES" if the employee signed the document package and "NO" otherwise. Print all the words without the quotes.

Sample test(s)
input
4 9
1 4 3
2 4
3 3 1
1 2 3
2 2
3 1 2
1 3 1
2 2
3 1 3
output
YES
NO
YES

题意：

给出 N 个结点和 M 个事件，1 代表 x 是 y 的 boss，2 代表从 x 开始往上传文件，3 代表询问第 x 个人有没有收到过这个文件。每次往上传文件都是传到最高的 boss 为止。输出每次询问的结果，有则输出 YES，没有则输出 NO。

思路：

并查集。维护两个并查集，一个是更新的，一个是不更新的，并且记录每次文件从哪里开始到哪里结束。最后询问的时候，查找不更新的并查集即可。

AC：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int NMAX = 100005;

int root[NMAX], num[NMAX];
int f[NMAX], tt[NMAX];

int Find (int x) {
    return x == root[x] ? x : root[x] = Find(root[x]);
}

int main() {

    int n, m;
    scanf("%d%d", &n, &m);

    for (int i = 1; i <= n; ++i) {
        root[i] = num[i] = i;
    }

    int ans = 0;
    while (m--) {
        int t;
        scanf("%d", &t);

        if (t == 1) {
            int x, y;
            scanf("%d%d", &x, &y);
            num[x] = root[x] = y;
        } else if (t == 2) {
            int x;
            scanf("%d", &x);
            ++ans;
            f[ans] = x;
            tt[ans] = Find(x);
        } else if (t == 3) {

            int x, i;
            scanf("%d%d", &x, &i);

            bool temp = false;
            int k = f[i];
            if (k == x || x == tt[i]) temp = true;

            while (!temp && k != tt[i]) {
                if (k == x) {
                    temp = true;
                    break;
                } else k = num[k];
            }

            if (temp) printf("YES\n");
            else printf("NO\n");

        }

    }

    return 0;
}