并查集区间合并
有一段序列从1到n，现在要把从x到y区间内的点合并起来。最简单暴力的方法就是抓住第一个去遍历剩下的(y-x)个点，或者抓住y去遍历x到(y-1)的点来合并，显然这样非常浪费时间。
对一个区间内的点进行合并，可以先用一个nextt[i]数组来标记从i点向右到第一个还没有合并的点的位置，初始化nextt[i]=i+1，是指一开始每个点都是孤立的，所以都指向下一个点（向右第一个还没有合并的点就是i+1）。这样在操作遍历区间的时候可以利用这个数组来直接跳到向右数第一个没有合并的点，然后去和这个点来合并。
Codeforces 556D Restructuring Company
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 some department. 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 use team(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 . 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.
Examples
Input
8 6
3 2 5
1 2 5
3 2 5
2 4 7
2 1 2
3 1 7
Output
NO
YES
YES
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <map>
#include <string>
#include <algorithm>
using namespace std;
int father[200005], nextt[200005];
int find(int x)
{
if (x==father[x])
return father[x];
return father[x]=find(father[x]);
}
void join(int x, int y)
{
int fx=find(x);
int fy=find(y);
if (fx!=fy)
father[fy]=fx;
}
int main()
{
int n, m;
scanf("%d%d", &n, &m);
for (int i=1; i<=n; i++){
father[i]=i;
nextt[i]=i+1;
}
int op, x, y;
while (m--){
scanf("%d%d%d", &op, &x, &y);
if (op==1)
join(x, y);
else if (op==2){
int temp;
for (int i=x+1; i<=y; i=temp){
join(i-1, i);
temp=nextt[i];//下一步让i直接跳到向右数第一个还没合并的点
nextt[i]=y+1;//因为把这个区间的所有点都合并后，所有点向右第一个还没合并的点都成了y+1
}
}else{
if (find(x)==find(y))
printf("YES\n");
else
printf("NO\n");
}
}
return 0;
}