BZOJ_3048_[Usaco2013 Jan]Cow Lineup

BZOJ_3048_[Usaco2013 Jan]Cow Lineup _双指针

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 bl

如果某个区间内数的种类不超过(k+1)种，那么这个区间的答案就是出现最多次数的数。

延伸一下，当左端点固定时，只需要求最长的一个区间使得数的种类不超过(k+1)种。

于是双指针法扫一遍数组，每次用当前数出现的次数来更新答案即可。

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define N 100050
int h[N];
struct A {
    int num,v,id;
}a[N];
int n,K;
bool cmp1(const A &x,const A &y) {
    return x.num<y.num;
}
bool cmp2(const A &x,const A &y) {return x.id<y.id;}
int main() {
    scanf("%d%d",&n,&K);
    int i,j;a[0].num=3423424;
    for(i=1;i<=n;i++) {
        scanf("%d",&a[i].num); a[i].id=i;
    }
    sort(a+1,a+n+1,cmp1);
    for(j=0,i=1;i<=n;i++) {
        if(a[i].num!=a[i-1].num)j++;
        a[i].v=j;
    }
    sort(a+1,a+n+1,cmp2);
    int tot=0,ans=0;
    int l=1,r=1;
    while(1) {
        while(r<=n&&tot<=K+1) {
            h[a[r].v]++; if(h[a[r].v]==1) tot++; ans=max(ans,h[a[r].v]); r++;
        }
        if(r==n+1) break;
        while(tot>K+1) {
            h[a[l].v]--; if(h[a[l].v]==0) tot--; l++;
        }
    }
    printf("%d
",ans);
}