bzoj 3048[Usaco2013 Jan]Cow Lineup 思想，乱搞 stl

3048: [Usaco2013 Jan]Cow Lineup

Time Limit: 2 Sec  Memory Limit: 128 MB
Submit: 237  Solved: 168
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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

这题刚开始不会，后来才知道在一个序列中，必须是连续，然后只有k+1种数，然后找出从中的

出现过的数的次数的最大值即可，用map维护，每次直接更新当前加入后该元素个数，不需要找最大值

因为在答案中，绝对是以其结尾出现过的。

 #pragma GCC optimize(2)
 #pragma G++ optimize(2)
 #include<iostream>
 #include<algorithm>
 #include<cmath>
 #include<cstdio>
 #include<cstring>
 #include<map>

 #define N 10000007
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(!isdigit(ch)){if(ch=='-')f=-;ch=getchar();}
     while(isdigit(ch)){x=(x<<)+(x<<)+ch-'';ch=getchar();}
     return x*f;
 }

 int n,k,ans;
 int a[N];
 map<int,int>p;

 int main()
 {
     n=read(),k=read()+;
     for (int i=;i<=n;i++)a[i]=read();
     int num=,mx=,l=;
     for (int i=;i<=n;i++)
     {
         if(!p[a[i]])num++;
         p[a[i]]++;
         while(num>k)
         {
             if(p[a[l]]==)p.erase(a[l]),num--;
             else p[a[l]]--;
             l++;
         }
         ans=max(ans,p[a[i]]);
 //        cout<<i<<" "<<ans<<endl;
     }
     printf("%d\n",ans);
 }