Educational Codeforces Round 7 C. Not Equal on a Segment 并查集

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; }