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;
}
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