HDU

Avin is studying series. A series is called "wave" if the following conditions are satisfied: 
1) It contains at least two elements; 
2) All elements at odd positions are the same; 
3) All elements at even positions are the same; 
4) Elements at odd positions are NOT the same as the elements at even positions. 
You are given a series with length n. Avin asks you to find the longest "wave" subseries. A subseries is a subsequence of a series.

InputThe first line contains two numbers n, c (1 ≤ n ≤ 100, 000, 1 ≤ c ≤ 100). The second line contains n integers whose range is [1, c], which represents the series. It is guaranteed that there is always a "wave" subseries.OutputPrint the length of the longest "wave" subseries.Sample Input

5 3
1 2 1 3 2

Sample Output

4

把任意两个数的位置存一下 然后暴力枚举进行依次存放即可
代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>

const int maxn=1e5+5;
typedef long long ll;
using namespace std;
vector<int>vec[105];
int main()
{
    int n,c;
    cin>>n>>c;
    int x;
    for(int t=1;t<=n;t++)
    {
        scanf("%d",&x);
        vec[x].push_back(t);
    }
    int ans=0;
    for(int t=1;t<=c;t++)
    {
        for(int j=1;j<=c;j++)
        {
            if(t==j)continue;
            int s1=vec[t].size();
            int s2=vec[j].size();
            int st1=0,st2=0;
            int temp=0;
            int sum=0;
            for(;;)
            {
                while(st1<s1&&vec[t][st1]<temp)
                {
                    st1++;
                }
                if(st1==s1)
                {
                    break;
                }
                temp=vec[t][st1];
                sum++;
                while(st2<s2&&vec[j][st2]<temp)
                {
                    st2++;
                }
                if(st2==s2)
                {
                    break;
                }
                sum++;
                temp=vec[j][st2];
            }
            ans=max(ans,sum);
        }
    }
    printf("%d
",ans);
    return 0;
}