2019CCPC-江西省赛（重现赛）D - Wave HDU - 6570

题文：

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

再给定序列中找满足条件的最长子序列，子序列要求最少两个元素，所有偶数下标位上元素都相同，所有奇数下标位上元素都相同。

感想

当时知道是DP，可是就是找不到思路，过后上网找题解也弄不懂大神们的思路，后来想了两天终于想出思路A了，终归还是自己能力不足，要再加把劲。

思路

1：DP的题目，切入点是问题的初始状态，这道题的初始状态显然为只有两个元素的时候（1个元素时不符合条件），这里设为（1， 2）

2：假设第三个数b为新的数，那么b不能和前面（1，2）构成更长的子序列，只能构成两个新子序列（1，b）和（2，b）

3：当数b为1时，可以延长（1，2），得到dp(1，2，1) = dp(1，2) + 1。由于所求子序列为重复的两个元素不断翻转，故而得出状态转移方程应为：
dp(a，b) = dp(b，a) + 1

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[110][110];
int book[110];
int main(){
    int n, c;while(scanf("%d%d", &n, &c) == 2){
    memset(dp, -1, sizeof(dp));memset(book, 0, sizeof(book));int num, len = 0;for(int i = 0; i < n; i++){
    scanf("%d", &num);book[num] = 1;//记录前面出现过的数for(int j = 1; j <= c; j++){
    if(j == num || book[j] == 0) continue;if(dp[num][j] == -1)//前面没有翻转的子序列dp[j][num] = 2;//初始化为2elsedp[j][num] = dp[num][j] + 1;len = max(len, dp[j][num]);}}printf("%d\n", len);}
}