Educational Codeforces Round 7 C. Not Equal on a Segment 并查集
C. Not Equal on a Segment
题目连接:
http://www.codeforces.com/contest/622/problem/C
Description
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.
Sample Input
6 4
1 2 1 1 3 5
1 4 1
2 6 2
3 4 1
3 4 2
Sample Output
2
6
-1
4
Hint
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
题解:
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
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+7;
int a[maxn];
int fa[maxn];
int fi(int x)
{
return x == fa[x]?x:fa[x]=fi(fa[x]);
}
int uni(int x,int y)
{
int p = fi(x),q = fi(y);
if(p != q)
{
fa[p]=fa[q];
}
}
int main()
{
int n,m;
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])
uni(i,i-1);
}
for(int i=1;i<=m;i++)
{
int flag = 0;
int l,r,x;
scanf("%d%d%d",&l,&r,&x);
int p = r;
while(p>=l)
{
if(a[p]!=x)
{
printf("%d\n",p);
flag = 1;
break;
}
p=fi(p)-1;
}
if(flag==0)printf("-1\n");
}
}
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