題目鏈接

題意

求至少包含$K$個給定字符串長度為$N$的字符串

思路

把所有可能的字符串建AC自動機，遍歷所有節點$dp[i][j][k]$ 表示以節點$j$為終點的長度為$i$包含狀態為k的字符串的方案數，$dp[i][j][k]=dp[i][j][k]+dp[i?1][fa[j]][t],t\in (0,1<<m)$

#include <bits/stdc++.h> const int maxn = 1e5 + 5; const int mod = 20090717; const int inf = 0x3f3f3f3f; using namespace std; struct Trie{int nex[maxn][26], fail[maxn], end[maxn];int root, p;inline int newnode() {for (int i = 0; i < 26; ++i) {nex[p][i] = -1;}end[p++] = 0;return p - 1;}inline void init() {p = 0;root = newnode();}inline void insert(char *buf, int id) {int now = root;for (int i = 0; buf[i]; ++i) {if (nex[now][buf[i]-'a'] == -1) nex[now][buf[i]-'a'] = newnode();now = nex[now][buf[i]-'a'];}end[now] |= (1 << id);} inline void build() {queue<int> que;fail[root] = root;for (int i = 0; i < 26; ++i) {if (nex[root][i] == -1)nex[root][i] = root;else {fail[nex[root][i]] = root;que.push(nex[root][i]);}}while (!que.empty()) {int now = que.front();que.pop();end[now] |= end[fail[now]];for (int i = 0; i < 26; ++i) {if (nex[now][i] == -1) nex[now][i] = nex[fail[now]][i];else {fail[nex[now][i]] = nex[fail[now]][i];que.push(nex[now][i]);}}}}}ac; char s[15]; int dp[26][101][1025]; int num(int x) {int cnt = 0;while (x) cnt += x & 1, x >>= 1;return cnt; } int main() {ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int n, m, p;while (scanf("%d %d %d", &n, &m, &p), n) {ac.init();for (int i = 0; i < m; ++i) {scanf("%s", s);ac.insert(s, i);}ac.build();memset(dp, 0, sizeof(dp));dp[0][0][0] = 1;for (int i = 1; i <= n; ++i) {for (int j = 0; j < ac.p; ++j) {for (int q = 0; q < (1 << m); ++q) {if (dp[i-1][j][q] == 0) continue;for (int k = 0; k < 26; ++k) {int v = ac.nex[j][k];dp[i][v][q | ac.end[v]] = (dp[i][v][q | ac.end[v]] + dp[i-1][j][q]) % mod;}}}}int ans = 0;for (int i = 0; i < ac.p; ++i) {for (int j = 0; j < (1 << m); ++j) {if (num(j) < p) continue;ans = (ans + dp[n][i][j]) % mod;}}printf("%d\n", ans);}return 0; }

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