D. Longest k-Good Segment

D. Longest k-Good Segment

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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/printfinstead 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 test(s)

input

5 5
1 2 3 4 5

output

1 5

input

9 3
6 5 1 2 3 2 1 4 5

output

3 7

input

3 1
1 2 3

output

1 1

two pointers一種不錯的編程技巧。

從前往後，或者從兩頭向中間，可使算法時間複雜度大大降低。

題目要求從所給序列找到一段儘量長的連續系列，使得這段序列不同數字個數(cnt)不大於K。

一個暴力的思路是：對於原始序列枚舉每個點作爲左端點，找到對應最大的右端點。記爲(Li,Ri)

假設從左向右邊枚舉，當枚舉Li+1時，先處理Li位置，然後，明顯當前Ri<=Ri+1;從Ri+1開始修改cnt值就可以了。

最後取最大值（Ri-Li）。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<set>
#include<vector>
#include<time.h>
#include<string.h>
using namespace std;
#define LL __int64
#define N 1000005
int a[N],num[N];

int main()
{
    int i,n,k;
    int l,r,cnt,pos;
    while(~scanf("%d%d",&n,&k)){
        for(i=1;i<=n;++i){
            scanf("%d",&a[i]);
        }
        l=r=1;
        cnt=1;
        pos=2;
        memset(num,0,sizeof(num));
        num[a[1]]=1;
        for(i=1;i<=n;++i){
            while(pos<=n&&cnt<=k){
                ++num[a[pos]];
                if(num[a[pos]]==1){
                    ++cnt;
                }
                if(cnt<=k)
                    ++pos;
            }
            if(r-l<pos-1-i){
                r=pos-1;
                l=i;
            }
            --num[a[i]];
            if(num[a[i]]==0){
                --cnt;
                ++pos;
            }
        }
        printf("%d %d\n",l,r);
    }
    return 0;
}