C. Not Equal on a Segment
time limit per test 1 second
memory limit per test 256 megabytes
input standard input
output standard output
You are given array a with n integers and m queries. The i-th query is given with three integers li,?ri,?xi.For the i-th query find any position pi (li?≤?pi?≤?ri) so that api?≠?xi.Input
The first line contains two integers n,?m (1?≤?n,?m?≤?2·105) — the number of elements in a and the number of queries.The second line contains n integers ai (1?≤?ai?≤?106) — the elements of the array a.Each of the next m lines contains three integers li,?ri,?xi (1?≤?li?≤?ri?≤?n,?1?≤?xi?≤?106) — the parameters of the i-th query.Output
Print m lines. On the i-th line print integer pi — the position of any number not equal to xi in segment [li,?ri] or the value ?-?1 if there is no such number.Examples
input
6 4
1 2 1 1 3 5
1 4 1
2 6 2
3 4 1
3 4 2
output
2
6
-1
4
题意：有n个数，然后m个询问，每次询问给你l,r,x，让你找到一个位置k，使得a[k]!=x,且l<=k<=r
思路：并查集，我们将a[i]=a[i-1]的合并在一起
#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=1e6+5;
int a[maxn];
int fa[maxn];
int Find(int x)
{return x==fa[x]?x:fa[x]=Find(fa[x]);
}
void un(int x,int y)
{int p=Find(x);int q=Find(y);if(p!=q)fa[p]=fa[q];
}int main()
{int n,m,i,flag=0;int l,r,x,p;scanf("%d%d",&n,&m);for(i=1;i<=n;i++)fa[i]=i;for(i=1;i<=n;i++){scanf("%d",&a[i]);if(a[i]==a[i-1])un(i,i-1);}while(m--){flag=0;scanf("%d%d%d",&l,&r,&x);p=r;while(p>=l){if(a[p]!=x){printf("%d\n",p);flag=1;break;}p=Find(p)-1;}if(flag==0)printf("-1\n");}return 0;
}