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 51 2 3 4 5

output

1 5

input

9 36 5 1 2 3 2 1 4 5

output

3 7

input

3 11 2 3

output

1 1

two pointers一种不错的编程技巧。

从前往后，或者从两头向中间，可使算法时间复杂度大大降低。

题目要求从所给序列找到一段尽量长的连续系列，使得这段序列不同数字个数(cnt)不大于K。

一个暴力的思路是：对于原始序列枚举每个点作为左端点，找到对应最大的右端点。记为(Li,Ri)

假设从左向右边枚举，当枚举Li+1时，先处理Li位置，然后，明显当前Ri<=Ri+1;从Ri+1开始修改cnt值就可以了。

最后取最大值（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 1000005int 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;}