Codeforces 813D Two Melodies dp

dp[ i ][ j ] 表示当前到处理完前 i 个，其中一个子串最后一位时 i ，另一个子串最后一位是 j 的最大值。
随便维护一下就能转移了。
#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 5000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, a[N], b[N];
int maxdp1[N][N];
int maxdp2[N][7];
int dp[N][N];

int hs[N], tot;
int toS[N], toB[N];

int main() {
    memset(maxdp1, 0xc0, sizeof(maxdp1));
    memset(maxdp2, 0xc0, sizeof(maxdp2));
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        hs[++tot] = a[i];
    }
    sort(hs + 1, hs + 1 + tot);
    tot = unique(hs + 1, hs + 1 + tot) - hs - 1;
    for(int i = 1; i <= n; i++) {
        b[i] = lower_bound(hs + 1, hs + 1 + tot, a[i]) - hs;
        if(binary_search(hs + 1, hs + 1 + tot, a[i] - 1)) toS[i] = b[i] - 1;
        else toS[i] = -1;
        if(binary_search(hs + 1, hs + 1 + tot, a[i] + 1)) toB[i] = b[i] + 1;
        else toB[i] = -1;
    }
    int ans = 0;
    for(int i = 1; i <= n; i++) {
        dp[i][0] = 1;
        if(~toS[i]) chkmax(dp[i][0], maxdp1[0][toS[i]] + 1);
        if(~toB[i]) chkmax(dp[i][0], maxdp1[0][toB[i]] + 1);
        chkmax(dp[i][0], maxdp2[0][a[i] % 7] + 1);
        for(int j = 1; j < i; j++) {
            dp[i][j] = 2;
            if(~toB[i]) chkmax(dp[i][j], maxdp1[j][toB[i]] + 1);
            if(~toS[i]) chkmax(dp[i][j], maxdp1[j][toS[i]] + 1);
            chkmax(dp[i][j], maxdp2[j][a[i] % 7] + 1);
            chkmax(dp[i][j], dp[j][0] + 1);
            chkmax(ans, dp[i][j]);
        }
        chkmax(maxdp1[0][b[i]], dp[i][0]);
        chkmax(maxdp2[0][a[i] % 7], dp[i][0]);
        for(int j = 1; j < i; j++) {
            chkmax(maxdp1[j][b[i]], dp[i][j]);
            chkmax(maxdp2[j][a[i] % 7], dp[i][j]);
            chkmax(maxdp1[i][b[j]], dp[i][j]);
            chkmax(maxdp2[i][a[j] % 7], dp[i][j]);
        }
    }
    printf("%d
", ans);
    return 0;
}

/*
*/