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。
解法： 并查集，我们将a[i]=a[i-1]的合并在一起，这样，我们就能一下子跳很多了
由于是找到不相等的，所以最多跳一步就能出结果，所以复杂度应该是比nlogn小的。

//CF 622C
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6+7;
int n, m, a[maxn];
namespace dsu{
    int fa[maxn];
    inline void init(){for(int i = 1; i <= n; i++) fa[i] = i;}
    inline int find_set(int x){if(x == fa[x]) return x; else return fa[x] = find_set(fa[x]);}
    inline void union_set(int x, int y){x = find_set(x), y = find_set(y); if(x != y) fa[x] = y;}
}
using namespace dsu;

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) fa[i] = i;
    for(int i = 1; i <= n; i++){
        scanf("%d", &a[i]);
        if(a[i] == a[i-1]) union_set(i, i-1);
    }
    for(int i = 1; i <= m; i++){
        int l, r, x;
        scanf("%d%d%d", &l, &r, &x);
        int p = r;
        bool ok = 0;
        while(p >= l){
            if(a[p] != x){
                printf("%d\n", p);
                ok = 1;
                break;
            }
            p = find_set(p) - 1;
        }
        if(ok == 0) puts("-1");
    }
    return 0;
}