A - Longest k-Good Segment （尺取，set， map）

最新推荐文章于 2022-03-05 17:07:08 发布

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本文介绍了一个算法问题，即如何找到一个数组中最长的k-好区间，该区间内的不同元素数量不超过k个。通过使用滑动窗口的方法，有效地解决了这一问题。

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.

Example
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

大意：

首先，真的很开心，尺取法学了一晚上，自己终于敲出来一个题目了！！
题意是找出最长的能够满足有不大于k个不同的字符的区间

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
using namespace std;
int a[12311324];
set <int>st;//记录有几个不同的数
map <int, int>cnt;//是否出现
int main()
{
    int n, m;
    scanf("%d %d", &n, &m);
    for(int i=1; i<=n; i++)
    {
        scanf("%d", a+i);
        st.insert(a[i]);
    }
    int num = st.size();
    int en = 1, st = 1, l = 1, r = 1, ka = 0;
    int sum = 0;
    if(num<=m)//所有都满足
        printf("%d %d\n", 1, n);
    else
    {
        while(en<=n)
        {
           if(sum<=m)//向右
           {
               if(cnt[a[en++]]++==0)
                sum++;
           }
           if(sum>m)//向左
           {
               if(--cnt[a[st++]]==0)
                sum--;
           }
            if(sum<=m)//更新
            {
                if(ka<en-st)
                {
                    ka = en - st;
                    l = st;
                    r = en-1;
                }
            }
        }
        printf("%d %d\n", l, r);
    }
    return 0;
}