D. Restructuring Company 并查集 +　维护一个区间技巧

http://codeforces.com/contest/566/problem/D

Even the most successful company can go through a crisis period when you have to make a hard decision — to restructure, discard and merge departments, fire employees and do other unpleasant stuff. Let's consider the following model of a company.

There are n people working for the Large Software Company. Each person belongs to somedepartment. Initially, each person works on his own project in his own department (thus, each company initially consists of n departments, one person in each).

However, harsh times have come to the company and the management had to hire a crisis manager who would rebuild the working process in order to boost efficiency. Let's useteam(person) to represent a team where person person works. A crisis manager can make decisions of two types:

Merge departments team(x) and team(y) into one large department containing all the employees of team(x) and team(y), where x and y (1 ≤ x, y ≤ n) — are numbers of two of some company employees. If team(x) matches team(y), then nothing happens.
Merge departments team(x), team(x + 1), ..., team(y), where x and y (1 ≤ x ≤ y ≤ n) — the numbers of some two employees of the company.

At that the crisis manager can sometimes wonder whether employees x and y (1 ≤ x, y ≤ n) work at the same department.

Help the crisis manager and answer all of his queries.

Input

The first line of the input contains two integers n and q (1 ≤ n ≤ 200 000, 1 ≤ q ≤ 500 000) — the number of the employees of the company and the number of queries the crisis manager has.

Next q lines contain the queries of the crisis manager. Each query looks like type x y, where $\text{[math]}$ . If type = 1 or type = 2, then the query represents the decision of a crisis manager about merging departments of the first and second types respectively. If type = 3, then your task is to determine whether employees x and y work at the same department. Note that x can be equal to y in the query of any type.

Output

For each question of type 3 print "YES" or "NO" (without the quotes), depending on whether the corresponding people work in the same department.

最麻烦就是第2种情况了，合并一个区间。

但是合并一次过后，它就是属于同一个组了，那么可以设tonext[i]表示第i个人所属的组，的下一个组，就是下一个和它合并的组。

开始的时候tonext[i] = i + 1

那么如果合并[x, y]区间，首先合并x和tonext[x]，然后合并tonext[x] 和 tonext[tonext[x]]

那么回溯的时候顺便更新个tonext，以后合并就是跳着合并了，不会一个个暴力合并。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 200000 + 20;
int fa[maxn];
int tonext[maxn];
int tofind(int x) {
    if (fa[x] == x) return x;
    else return fa[x] = tofind(fa[x]);
}
void tomerge(int u, int v) {
    u = tofind(u);
    v = tofind(v);
    fa[v] = u;
}

void dfs(int be, int en, int cur, int pre) {
    if (cur > en) {
        tonext[pre] = cur;
        return;
    }
    tomerge(pre, cur);
    dfs(be, en, tonext[cur], cur);
    tonext[pre] = tonext[cur];
}

void work() {
    int n, q;
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; ++i) {
        fa[i] = i;
        tonext[i] = i + 1;
    }
    for (int i = 1; i <= q; ++i) {
        int op, x, y;
        scanf("%d%d%d", &op, &x, &y);
        if (op == 1) {
            tomerge(x, y);
        } else if (op == 2) {
            dfs(x, y, tonext[x], x);
        } else {
            if (tofind(x) == tofind(y)) {
                printf("YES
");
            } else printf("NO
");
        }
    }
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return 0;
}

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D. Restructuring Company 并查集 + 维护一个区间技巧

D. Restructuring Company 并查集 +　维护一个区间技巧