2019杭電0場ccpc江西省賽（重現）Wave

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
題意：
題目給你一個序列，找一個波的最大長度，波定義爲：
1）它包含至少兩個元素;
2）奇數位置的所有元素都是相同的;
3）偶數位置的所有元素都是相同的;
4）奇數位置的元素與偶數位置的元素不同
這個題比賽時亨亨寫了一發優化暴力直接a了，後來聽學長講解，第一次聽到序列自動機這個高bi格的東西，即一個二維的數組next [i][j]，表示在數i的第j個的位置，然後枚舉奇偶兩個位置，保存最大值；
預處理：

for(int i=1;i<=n;i++)
    {
        ll k;
        scanf("%lld",&k);
        a[k].push_back(i);			//將輸入的每個數的位置保存下來，之後就可以遍歷奇數和偶數的位置；
    }

遍歷

for(int i=1;i<=c;i++)
    {
        for(int j=1;j<=c;j++)
        {
            if(i==j) continue;
            ll num=0;
            ll s1=a[i].size(),s2=a[j].size();
            //printf("%d: %lld %d: %lld\n",i,s1,j,s2);
            ll k=0,p=0,q=0;
            while(1)
            {
                while(p<s1&&k>a[i][p])                     //遍歷奇數
                //用k記錄上一個遍歷的位置，下一個位置要>K才行；
                    p++;
                if(p==s1)
                    break;
                num++;
                k=a[i][p];
                while(q<s2&&k>a[j][q])                      //遍歷偶數
                    q++;
                if(q==s2)
                    break;
                k=a[j][q];
                num++;
            }
            ans=max(ans,num);
            //printf("%lld\n",ans);
        }
    }

代碼

#include<bits/stdc++.h>
#include<map>
#define ll long long
using namespace std;
const int maxn=1e5+7;
const int INF=1e9;
vector<ll >a[105];
int main()
{
    ll n,c;
    scanf("%lld%lld",&n,&c);
    for(int i=1;i<=n;i++)
    {
        ll k;
        scanf("%lld",&k);
        a[k].push_back(i);
    }
    /*for(int i=1;i<=c;i++)
    {
        printf("%d: ",i);
        for(int j=0;j<a[i].size();j++)
            printf("%lld ",a[i][j]);
        printf("\n");
    }*/
    ll ans=0;
    for(int i=1;i<=c;i++)
    {
        for(int j=1;j<=c;j++)
        {
            if(i==j) continue;
            ll num=0;
            ll s1=a[i].size(),s2=a[j].size();
            //printf("%d: %lld %d: %lld\n",i,s1,j,s2);
            ll k=0,p=0,q=0;
            while(1)
            {
                while(p<s1&&k>a[i][p])                     //奇數
                    p++;
                if(p==s1)
                    break;
                num++;
                k=a[i][p];
                while(q<s2&&k>a[j][q])                      //偶數
                    q++;
                if(q==s2)
                    break;
                k=a[j][q];
                num++;
            }
            ans=max(ans,num);
            //printf("%lld\n",ans);
        }
    }
    printf("%lld\n",ans);
    return 0;
}