【CF813D】Two Melodies

【CF813D】Two Melodies

题面

洛谷

题解

$dp$:

设$f[i][j]$表示第一个集合以$i$结尾、第二个集合以$j$结尾的合法长度之和最大是多少

明显有$f[i][j]=f[j][i]$

所以不妨设$i<j$

暴力就是$O(n^3)$的

然后因为合法的转移只有它的绝对值相差一或模$7$

所以直接开两个桶维护一下

模$7$最大的$dp$值、值为所有定值为$dp$值即可

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (!isdigit(ch) && ch != '-') ch = getchar();
    if (ch == '-') w = -1, ch = getchar();
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
    return w * data;
}
const int MAX_N = 5e3 + 5;
const int MAX_M = 1e5 + 5;
int N, a[MAX_N], f[MAX_N][MAX_N], bln_nxt[MAX_M], bln_mod[10];
int ans = 0;
int main () {
    N = gi(); for (int i = 1; i <= N; i++) a[i] = gi();
    for (int i = 0; i <= N; i++) {
        memset(bln_mod, 0, sizeof(bln_mod));
        memset(bln_nxt, 0, sizeof(bln_nxt));
        for (int j = 1; j < i; j++) {
            bln_nxt[a[j]] = max(bln_nxt[a[j]], f[i][j]);
            bln_mod[a[j] % 7] = max(bln_mod[a[j] % 7], f[i][j]);
        }
        for (int j = i + 1; j <= N; j++) {
            f[i][j] = max(bln_nxt[a[j] - 1], bln_nxt[a[j] + 1]) + 1;
            f[i][j] = max(f[i][j], f[i][0] + 1);
            f[i][j] = max(f[i][j], bln_mod[a[j] % 7] + 1);
            f[j][i] = f[i][j];
            bln_nxt[a[j]] = max(bln_nxt[a[j]], f[i][j]);
            bln_mod[a[j] % 7] = max(bln_mod[a[j] % 7], f[i][j]);
            ans = max(ans, f[i][j]);
        }
    }
    printf("%d\n", ans);
    return 0;
}

参考文章