Codeforces Round #116 E-Cubes 180E_let's imagine that you're playing the following si-优快云博客

本文链接：https://blog.youkuaiyun.com/sunny606/article/details/7489237

本文介绍了一款名为E.Cubes的游戏中的最优解法，玩家需删除不超过k个方块使连续相同颜色的方块序列最长。文章提供了一个O(n)算法实现，通过维护颜色区间来快速计算最大得分。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Let's imagine that you're playing the following simple computer game. The screen displays n lined-up cubes. Each cube is painted one ofm colors. You are allowed to delete not more than k cubes (that do not necessarily go one after another). After that, the remaining cubes join together (so that the gaps are closed) and the system counts the score. The number of points you score equals to the length of the maximum sequence of cubes of the same color that follow consecutively. Write a program that determines the maximum possible number of points you can score.

Remember, you may delete no more than k any cubes. It is allowed not to delete cubes at all.

Input

The first line contains three integers n, m and k (1 ≤ n ≤ 2· 10⁵, 1 ≤ m ≤ 10⁵, 0 ≤ k < n). The second line contains n integers from 1to m — the numbers of cube colors. The numbers of colors are separated by single spaces.

Output

Print the maximum possible number of points you can score.

Sample test(s)

     input
   
10 3 2
1 2 1 1 3 2 1 1 2 2

     output
   
4

因为数据量达到了10^5，所以要O（n）或是O(nlogn)级别的算法才能过，这道题要做的是对区间上信息的统计，所以用线段树是可以做的，不过感觉这个问题和曾经遇到过的maximum sum(http://poj.org/problem?id=2479)有些相似，所以可能会有类似的解法，事实证明确实有O(n)的算法，思想和maximum sum的思想类似，所以建议先去看看maximum sum。

代码：

#include<cstdio>
#define MAX 200001

int last[MAX],next[MAX],max[MAX],input[MAX],sum[MAX],rec[MAX];

int main()
{
    int n,m,k;
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=n;i++) scanf("%d",&input[i]);
    for(int i=1;i<=n;i++)
    {
         int color=input[i];
         if(input[i]!=input[i-1])
         {
              next[last[color]]=i;
              last[color]=i;
         }
         sum[color]++;
         rec[i]=sum[color];
    }
    for(int i=0;i<=m;i++) last[i]=0;
    for(int i=1;i<=n;i++) if(last[input[i]]==0) last[input[i]]=i;
    for(int i=1;i<=n;i++)
    {
         int color=input[i];
         int same_color;
         while(true)
         {
              int total_color=i-last[color]+1;
              same_color=rec[i]-rec[last[color]]+1;
              int dis=total_color-same_color;
              if(dis<=k) break;
              last[color]=next[last[color]];
         }
         if(same_color>max[color]) max[color]=same_color;
    }
    int ans=0;
    for(int i=1;i<=m;i++) if(max[i]>ans) ans=max[i];
    printf("%d\n",ans);
    return 0;
}