Codeforces Educational Codeforces Round 5 D. Longest k-Good Segment 尺取法

D. Longest k-Good Segment

题目连接：

http://www.codeforces.com/contest/616/problem/D

Description

The array a with n integers is given. Let's call the sequence of one or more consecutive elements in a segment. Also let's call the segment k-good if it contains no more than k different values.

Find any longest k-good segment.

As the input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

input

The first line contains two integers n, k (1 ≤ k ≤ n ≤ 5·105) — the number of elements in a and the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 106) — the elements of the array a.

Output

Print two integers l, r (1 ≤ l ≤ r ≤ n) — the index of the left and the index of the right ends of some k-good longest segment. If there are several longest segments you can print any of them. The elements in a are numbered from 1 to n from left to right.

Sample Input

5 5

1 2 3 4 5

Sample Output

1 5

Hint

题意

给你n个数，你需要找到一个最长的区间，使得这个区间里面不同的数小于等于k个

题解：

尺取法扫一遍就好了

代码

#include<bits/stdc++.h>
using namespace std;

#define maxn 1000005
int a[maxn];
int vis[maxn];
int main()
{
    int n,k;scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    int al=0,ar=0,ans=0,now=0,l=1;
    for(int i=1;i<=n;i++)
    {
        vis[a[i]]++;
        if(vis[a[i]]==1)now++;
        while(now>k)
        {
            vis[a[l]]--;
            if(vis[a[l]]==0)
                now--;
            l++;
        }
        if(i-l+1>=ar-al+1)
            ar=i,al=l;
    }
    cout<<al<<" "<<ar<<endl;
}