题目：https://vjudge.net/contest/240521#problem/C
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 xand 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
按照题目一样，不过1，直接将两个人合并，2就有点麻烦，因为如果你直接暴力就会t，从x-y有可能有的已经成为小团体了。因此我们需要重新想一种方法。利用一个数组维护记录每个人他后面的第一个不属于他部门的人的id。每次2操作时将这个区间的下一个都改成y的下一个。
#include <bits/stdc++.h>
#define MAXN 200005
using namespace std;
int pre[MAXN];
int next[MAXN];
int find(int x)
{
int r=x;
while(pre[r]!=r)
r=pre[r];
int i=x,j;//路径压缩
while(pre[i]!=r)
{
j=pre[i];
pre[i]=r;
i=j;
}
return r;
}
void Union(int x,int y)
{
int fx=find(x);
int fy=find(y);
if(fx!=fy)
pre[fy]=fx;
}
int main()
{
int n,p;
int op,x,y,t;
while(cin>>n>>p){
for(int i=1;i<=n;i++)
pre[i]=i,next[i]=i+1;
for(int i=1;i<=p;i++)
{
cin>>op>>x>>y;
if(op==1)
Union(x,y);
else if(op==2)
{
for(int i=x+1;i<=y;i=t)
{
//if(find(i-1)!=find(i))//如果没有合并过
Union(i-1,i);
t=next[i];
next[i]=next[y];
}
}
else
{
if(find(x)==find(y))cout<<"Yes";
else cout<<"No";
}
}
}
return 0;
}
不过还有一种做法，自己可以运行出来，不知道为什么提交上去就会WA，想法就是在执行2是判断是否属于同意个。
#include <bits/stdc++.h>
#define MAXN 200005
using namespace std;
int pre[MAXN];
int find(int x)
{
int r=x;
while(pre[r]!=r)
r=pre[r];
int i=x,j;//路径压缩
while(pre[i]!=r)
{
j=pre[i];
pre[i]=r;
i=j;
}
return r;
}
void Union(int x,int y)
{
int fx=find(x);
int fy=find(y);
if(fx!=fy)
pre[fy]=fx;
}
int main()
{
int n,p;
int op,x,y;
while(cin>>n>>p){
for(int i=1;i<=n;i++)
pre[i]=i;
for(int i=1;i<=p;i++)
{
cin>>op>>x>>y;
if(op==1)
Union(x,y);
else if(op==2)
{
for(int i=x+1;i<=y;i++)
{
if(find(i-1)!=find(i))//如果没有合并过
Union(i-1,i);
// t=next[i];
// next[i]=next[y];
}
}
else
{
if(find(x)==find(y))cout<<"Yes";
else cout<<"No";
}
}
}
return 0;
}