[hdu6570]Wave

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 262144/262144 K (Java/Others)

Problem Description
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.

Input
The 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.

Output
Print the length of the longest “wave” subseries.

Sample Input

5 3
1 2 1 3 2

Sample Output

4

题意：
给定一个数列，定义wave为：长度大于等于2的数列，且奇数位上只有一种数字，偶数位上也只有一种数字，奇数位数字跟偶数位数字不同。
题解：
DP
$f[i][j]$表示奇数位为 $i$,偶数位为 $j$的最长长度
顺着递推。
对于每个 $a[i]$,你可以更新所有 $f[a[i]][j]$为偶数的情况，和所有 $f[j][a[i]]$为奇数的情况。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n,c,a[100004];
int f[104][104];
int w33ha(){
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    memset(f,0,sizeof(f));
    for(int i=1;i<=n;i++){
        for(int j=1;j<=c;j++){
            if(a[i]==j)continue;
            if(f[a[i]][j]%2==0){
                f[a[i]][j]++;
            }
            if(f[j][a[i]]%2==1){
                f[j][a[i]]++;
            }
        }
    }
    int ans=0;
    for(int i=1;i<=c;i++){
        for(int j=1;j<=c;j++){
            ans=max(ans,f[i][j]);
        }
    }
    printf("%d\n",ans);
    return 0;
}
int main(){
    while(scanf("%d%d",&n,&c)!=EOF)w33ha();
    return 0;
}