bzoj3048[Usaco2013 Jan]Cow Lineup 尺取法

3048: [Usaco2013 Jan]Cow Lineup

Time Limit: 2 Sec  Memory Limit: 128 MB
Submit: 225  Solved: 159
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Description

 

 Farmer John's N cows (1 <= N <= 100,000) are lined up in a row. Each cow is identified by an integer "breed ID" in the range 0...1,000,000,000; the breed ID of the ith cow in the lineup is B(i). Multiple cows can share the same breed ID. FJ thinks that his line of cows will look much more impressive if there is a large contiguous block of cows that all have the same breed ID. In order to create such a block, FJ chooses up to K breed IDs and removes from his lineup all the cows having those IDs. Please help FJ figure out the length of the largest consecutive block of cows with the same breed ID that he can create by doing this.

 

给你一个长度为n(1<=n<=100,000)的自然数数列，其中每一个数都小于等于10亿，现在给你一个k，表示你最多可以删去k类数。数列中相同的数字被称为一类数。设该数列中满足所有的数字相等的连续子序列被叫做完美序列，你的任务就是通过删数使得该数列中的最长完美序列尽量长。

Input

* Line 1: Two space-separated integers: N and K. 
* Lines 2..1+N: Line i+1 contains the breed ID B(i). 

Output

* Line 1: The largest size of a contiguous block of cows with identical breed IDs that FJ can create.

 

Sample Input

9 1
2
7
3
7
7
3
7
5
7

INPUT DETAILS: There are 9 cows in the lineup, with breed IDs 2, 7, 3, 7, 7, 3, 7, 5, 7. FJ would like to remove up to 1 breed ID from this lineup.

Sample Output

4

OUTPUT DETAILS: By removing all cows with breed ID 3, the lineup reduces to 2, 7, 7, 7, 7, 5, 7. In this new lineup, there is a contiguous block of 4 cows with the same breed ID (7).

HINT

 

样例解释：

  长度为9的数列，最多只能删去1类数。

  不删，最长完美序列长度为2.

  删去一类数3，序列变成2 7 7 7 7 5 7，最长完美序列长度为4.因此答案为4.

 

Source

Gold

尺取法
记录区间内每个元素出现次数和颜色种数
当种数<=k+1时r++,当种数>k+1时l++

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#define N 100005
using namespace std;
int n,m,ans,num,a[N],b[N],cnt[N];
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    scanf("%d",&a[i]),b[i]=a[i];
    sort(b+1,b+1+n);
    int len=unique(b+1,b+1+n)-b-1;
    for(int i=1;i<=n;i++)
    a[i]=lower_bound(b+1,b+1+len,a[i])-b;
    int l=1,r=1;cnt[a[1]]++;num++;
    while(r<n){
        while(r<n&&num<=m+1){
            if(!cnt[a[++r]])num++;
            cnt[a[r]]++;
            ans=max(ans,cnt[a[r]]);
        }
        while(num>m+1&&l<=r)
        if(!(--cnt[a[l++]]))num--; 
    }
    printf("%d\n",ans);
    return 0;
}