CodeForces - 616D Longest k-Good Segment （尺取法）

最新推荐文章于 2022-05-20 11:17:19 发布

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本文介绍了一种解决寻找数组中最长k-好子段问题的方法，即子串中不同数值种类不超过k的最大长度。通过典型的尺取法实现，从数组的起始位置开始逐个元素扫描，同时维护一个计数器来跟踪当前子段内不同数值的数量。

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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.

题意：求子串中不同数的个数不超过k的最大长度

思路：典型的尺取法，从左往右一个一个爬，具体看代码

代码：

#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <numeric>
#include <set>
#include <string>
#include <cctype>
#include <sstream>
#define INF 0x3f3f3f3f
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
typedef long long LL;
using namespace std;
const int maxn = 5e5 + 5;

int n, k;
int a[maxn], vis[1000000 + 5], cnt = 0;
int main() {
    // freopen ("in.txt","r",stdin);
    scanf ("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) scanf ("%d", &a[i]);
    int ans = -1,anss,anst;
    int s = 1, t = 1;
    while (s<=n) {
        while (t <= n && cnt <= k) {
            if (!vis[a[t]]&&cnt==k) break;
            if (vis[a[t++]]++==0 ) {
                cnt++;
            }
        }
        if (t-s-1>ans){
            ans=t-s-1;
            anss=s;
            anst=t-1;
        }
        if (--vis[a[s++]]==0) cnt--;
    }
    printf ("%d %d\n",anss,anst);
    return 0;
}