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CF813D Two Melodies(dp)

題面

luogu
Codeforces

題目大意:

  • 給一個長度為\(n\)的序列,求兩個不相交的子集長度之和最大是多少,能放入同一子集的條件是首先順序不能變,然後每一個相鄰的要麼相差\(1\)或者相差\(7\)的倍數。

  • \(n<=5000\)

題解

\(dp:\)

\(f[i][j]\)表示第一序列到了第\(i\)位,第二個序列到了第\(j\)位,符合條件的長度之和最大

顯然, \(f[i][j] == f[j][i]\)

那麼我們可以只考慮\(i<j\)

暴力轉移是\(O(n^3)\)

顯然不行

注意 相鄰要麼相差\(1\),要麼相差\(7\)的倍數

對於相差\(1\), 開一個桶記錄\(max\)
相差\(7\),就是 模\(7\) 同餘

也開一個桶

這樣複雜度就是\(O(n^2)\)

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;
template<class T> inline void read(T &x) {
    x = 0; RG char c = getchar(); bool f = 0;
    while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
    while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
    x = f ? -x : x;
    return ;
}
template<class T> inline void write(T x) {
    if (!x) {putchar(48);return ;}
    if (x < 0) x = -x, putchar('-');
    int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
    for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
}
const int N = 5010, M = 100010;
int f[N][N], a[N];
int pre[M], mod[10];
int main() {
    int n, ans = 0; read(n);
    for (int i = 1; i <= n; i++) read(a[i]);
    for (int i = 0; i <= n; i++) {
        memset(mod, 0, sizeof(mod));
        memset(pre, 0, sizeof(pre));
        for (int j = 1; j < i; j++) {
            pre[a[j]] = max(pre[a[j]], f[i][j]);
            mod[a[j] % 7] = max(mod[a[j] % 7], f[i][j]);
        }
        for (int j = i+1; j <= n; j++) {
            f[i][j] = max(pre[a[j] - 1], pre[a[j] + 1]) + 1;
            f[i][j] = max(f[i][j], f[i][0] + 1);
            f[i][j] = max(f[i][j], mod[a[j] % 7] + 1);
            f[j][i] = f[i][j];
            pre[a[j]] = max(pre[a[j]], f[i][j]);
            mod[a[j] % 7] = max(mod[a[j] % 7], f[i][j]);
            ans = max(ans, f[i][j]);
        }
    }
    printf("%d\n", ans);
    return 0;
}