Codeforces-813D Two Melodies(dp)

D. Two Melodies
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Alice is a beginner composer and now she is ready to create another masterpiece. And not even the single one but two at the same time!
Alice has a sheet with n notes written on it. She wants to take two such non-empty non-intersecting subsequences that both of them form a melody and sum of their lengths is maximal.
Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Subsequence forms a melody when each two adjacent notes either differs by 1 or are congruent modulo 7.
You should write a program which will calculate maximum sum of lengths of such two non-empty non-intersecting subsequences that both of them form a melody.
Input
The first line contains one integer number n (2 ≤ n ≤ 5000).
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — notes written on a sheet.
Output
Print maximum sum of lengths of such two non-empty non-intersecting subsequences that both of them form a melody.
Examples
input

4
1 2 4 5

output

4

input

6
62 22 60 61 48 49

output

5

Note
In the first example subsequences [1, 2] and [4, 5] give length 4 in total.
In the second example subsequences [62, 48, 49] and [60, 61] give length 5 in total. If you choose subsequence [62, 61] in the first place then the second melody will have maximum length 2, that gives the result of 4, which is not maximal.
dp[i][j]를 두 개의 하위 서열의 끝으로 각각 i와 j의 최대 총산도를 설정하면 ij를 합쳐서 i를 열거할 수 있다. 그러면 현재 j는 이미 첫 번째 하위 서열에 의해 선택되었을 수도 있다.dp[i][j]와 dp[j][i]가 같기 때문에 매번 dp[i][j]를 업데이트할 때마다 dp[j]=dp[i][j]로 하여금 자연스럽게 i>j의 상황을 구할 수 있다.

#include 
#define x first
#define y second
using namespace std;
typedef long long LL;
const int MX = 5005;
const int MXM = 1e5 + 5;
int dp[MX][MX]; //dp[i][j]   i，j   2        
int a[MX], pre[MXM], mod[8];
int main() {
    //freopen("in.txt", "r", stdin);
    int n, ans = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 0; i <= n; i++) {  //      i，      j，       
        memset(pre, 0, sizeof(pre));
        memset(mod, 0, sizeof(mod));
        for (int j = 1; j < i; j++) {
            pre[a[j]] = max(pre[a[j]], dp[i][j]);//pre        a[j]      
            mod[a[j] % 7] = max(mod[a[j] % 7], dp[i][j]);//mod        7       
        }
        for (int j = i + 1; j <= n; j++) {
            dp[i][j] = max(pre[a[j] + 1], pre[a[j] - 1]) + 1;
            dp[i][j] = max(dp[i][j], mod[a[j] % 7] + 1);
            dp[i][j] = max(dp[i][j], dp[i][0] + 1);
            dp[j][i] = dp[i][j];    //  
            pre[a[j]] = max(pre[a[j]], dp[i][j]);
            mod[a[j] % 7] = max(mod[a[j] % 7], dp[i][j]);
            ans = max(ans, dp[i][j]);
        }
    }
    printf("%d
", ans);
    return 0;
}