UVA1401 Remember the Word 题解

考虑一个动态规划，$f_i$ 表示前 $i$ 个字符的划分方案。枚举 $i$，对于 $s$ 前 $i$ 个字符的某个后缀，如果出现在字典里，就转移。形式化地，如果 $s[j \cdots i]$ 为一个在字典的后缀，那么 $f_i = \sum \limits_{j} f_{j-1}$。$f_0=1$。

但直接转移是 $O(nm)$ 的。发现字典每个字符串长度不超过过 $100$，只需要枚举位数，用哈希判断即可。不过有点卡常。

#pragma GCC optimize("-Ofast,fast-math,-inline")
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <string>
#include <algorithm>
#include <unordered_set>
using namespace std;
using ull = unsigned long long;

const int N = 3e5 + 5, M = 4005, K = 105, MOD = 20071027;

long long f[N];
string s;
int n;
string t[M];

ull hashing[N], hash2[M][K];

ull powe[N];
unordered_set<ull> st;

int main()
{
	ios::sync_with_stdio(0), cin.tie(nullptr), cout.tie(nullptr);
	int c = 0;
	powe[0] = 1;
	for (int i = 1; i < N; i++)
	{
		powe[i] = powe[i - 1] * 841;
	}
	while (cin >> s)
	{
		st.clear();
		int p = s.size();
		for (int i = 1; i <= p; i++)
		{
			hashing[i] = hashing[i - 1] * 841 + s[i - 1];
		}
		cin >> n;
		for (int i = 1; i <= n; i++)
		{
			cin >> t[i];
			int pp = t[i].size();
			for (int j = 1; j <= pp; j++)
			{
				hash2[i][j] = hash2[i][j - 1] * 841 + t[i][j - 1];
			}
			st.insert(hash2[i][pp]);
		}
		f[0] = 1LL;
		for (int i = 1; i <= p; i++)
		{
			f[i] = 0LL;
			int maxn = min(i, 100);
			for (int j = 1; j <= maxn; j++)
			{
				int l = i - j + 1, r = i;
				ull subhash = hashing[r] - hashing[l - 1] * powe[r - l + 1];
				if (st.count(subhash)) f[i] = (f[i] + f[l - 1]) % MOD;
			}
		}
		cout << "Case " << ++c << ": " << f[p] << "\n";
	}
	return 0;
}