Codeforces 1129C Morse Code dp

Morse Code

每次加入一个字符的时候吧之前没有出现过的子串的方案数统计进去， 这个可以用字典树去check出现过没。

#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 = 3000 + 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;}
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
int m, n;
int ans;
int a[N];
int dp[N];
 
int tot = 1;
int ch[N * N][2];
 
bool check(int l, int r) {
    int x = a[l], y = a[l + 1], z = a[l + 2], w = a[l + 3];
    if(x == 0 && y == 0 && z == 1 && w == 1) return false;
    if(x == 0 && y == 1 && z == 0 && w == 1) return false;
    if(x == 1 && y == 1 && z == 1 && w == 0) return false;
    if(x == 1 && y == 1 && z == 1 && w == 1) return false;
    return true;
}
 
void calc(int n) {
    dp[n + 1] = 1;
    int u = 1;
    for(int i = n; i >= 1; i--) {
        dp[i] = 0;
        if(i + 1 <= n + 1) add(dp[i], dp[i + 1]);
        if(i + 2 <= n + 1) add(dp[i], dp[i + 2]);
        if(i + 3 <= n + 1) add(dp[i], dp[i + 3]);
        if(i + 4 <= n + 1 && check(i, i + 3)) add(dp[i], dp[i + 4]);
        if(!ch[u][a[i]]) {
            ch[u][a[i]] = ++tot;
            add(ans, dp[i]);
        }
        u = ch[u][a[i]];
    }
}
 
int main() {
    scanf("%d", &m);
    for(int i = 1; i <= m; i++) {
        scanf("%d", &a[i]);
    }
    for(int i = 1; i <= m; i++) {
        calc(i);
        printf("%d
", ans);
    }
    return 0;
}
 
/*
*/