USACO Cow Lineup

洛谷傳送門

JDOJ傳送門

Description

Problem 1: Cow Lineup [Brian Dean and Daniel Dara, 2012]

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.

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

Sample Output

4

題解：

允許刪K種，最後還必須連續，那麼想到滑動視窗。也就是，採用滑動視窗的模式遍歷所有包含種類數小於等於K+1的區間們。為什麼是K+1呢，因為最多刪除K種還剩一種。然後每次合法區間的最多種類來更新答案。

離散化開桶已經是套路的東西了，不需要說了。

然後我們思考如何快速找出每個合法區間的最多種類。

我們發現只需要拿cow[a[r]]來更新答案就行了。因為每次右移較之上次移動都只會有最右側的節點+。那麼每次更新也就是最右側節點。

程式碼：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<map>
using namespace std;
const int maxn = 100010;
int n,k,cow[maxn],a[maxn];
int cnt;
map<int,int> mp;
int getid(int num)
{
	if(!mp[num])mp[num]=++cnt;
	return mp[num];
}
int main()
{
	cin>>n>>k;
	int t;
	for(int i=1;i<=n;i++)
	{
		cin>>t;
		a[i]=getid(t);
	}
	int ans=1;
	int l=1,r=0,typ=0;
	while(r<=n)
	{
		r++;
		if(cow[a[r]]==0)
			typ++; 
		cow[a[r]]++;
		while(typ>k+1)
		{
			cow[a[l]]--;
			if(cow[a[l]]==0)
				typ--;
			l++;
		}
		ans=max(ans,cow[a[r]]); 
	}
	cout<<ans<<endl;
	return 0;
}