【Codeforces 622C】 Not Equal on a Segment （并查集思想）

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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;
}