BZOJ 1559: [JSOI2009]密码( AC自动机 + 状压dp )

建AC自动机后, dp(x, y, s)表示当前长度为x, 在结点y, 包括的串的状态为s的方案数, 转移就在自动机上走就行了. 对于输出方案, 必定是由给出的串组成(因为<=42), 所以直接暴搜答案. 数据范围很小, 可以AC(复杂度懒得算了....)

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#include<cstdio>

#include<iostream>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

#define b(x) (1 << (x))

#define C(c) (c - 'a')

const int maxn = 11;

int L, N, Com[maxn][maxn], lth[maxn], V[maxn], R[45], cnt;

ll dp[2][100][b(maxn)];

char S[maxn][maxn], res[45][30];

bool F[maxn];

struct Node {

Node *ch[26], *f;

int Id, n;

} pool[100], *pt = pool, *Root, *q[100];

void Init() {

scanf("%d%d", &L, &N);

(Root = pt++)->Id = 0;

int idn = Root->n = 0;

for(int i = 0; i < N; i++) {

scanf("%s", S[i]);

lth[i] = strlen(S[i]);

Node* t = Root;

for(int j = 0; j < lth[i]; j++) {

if(!t->ch[C(S[i][j])]) {

pt->Id = ++idn, pt->n = 0;

t->ch[C(S[i][j])] = pt++;

}

t = t->ch[C(S[i][j])];

}

t->n |= b(i);

}

for(int i = 0; i < N; i++)

for(int j = 0; j < N; j++) {

int &v = Com[i][j];

for(v = min(lth[i], lth[j]); v; v--) {

bool F = true;

int a = lth[i] - v, b = 0;

while(a < lth[i]) {

if(S[i][a] != S[j][b]) {

F = false;

break;

}

a++, b++;

}

if(F) break;

}

}

cnt = 0;

}

void Build() {

int h = 0, t = 0;

(q[t++] = Root)->f = NULL;

while(h < t) {

Node* o = q[h++];

for(int i = 0; i < 26; i++) if(o->ch[i]) {

Node* f = o->f;

while(f && !f->ch[i]) f = f->f;

o->ch[i]->f = f ? f->ch[i] : Root;

q[t++] = o->ch[i];

}

}

}

void DFS(int x, int len) {

if(x == N) {

if(len == L) {

int p = 0;

for(int i = 0; i < N; i++)

for(int j = i ? Com[V[i - 1]][V[i]] : 0; j < lth[V[i]]; j++)

res[cnt][p++] = S[V[i]][j];

cnt++;

}

return;

}

for(int i = 0; i < N; i++) if(!F[i]) {

F[V[x] = i] = true;

DFS(x + 1, len + lth[i] - Com[V[x - 1]][i]);

F[i] = false;

}

}

bool Cmp(const int &l, const int &r) {

return strcmp(res[l], res[r]) < 0;

}

void Work() {

int c = 0, p = 1, h = 0, t = 0, S = b(N) - 1;

q[t++] = Root;

while(h < t) {

Node* o = q[h++];

if(o->f) o->n |= o->f->n;

for(int i = 0; i < 26; i++)

if(o->ch[i]) q[t++] = o->ch[i];

}

memset(dp[c], 0, sizeof dp[c]);

dp[c][0][0] = 1;

for(int i = 0; i < L; i++) {

swap(c, p);

memset(dp[c], 0, sizeof dp[c]);

for(int r = 0; r < t; r++)

for(int s = b(N); s--; ) if(dp[p][q[r]->Id][s]) {

for(int j = 0; j < 26; j++) {

Node* o = q[r];

while(o && !o->ch[j]) o = o->f;

o = o ? o->ch[j] : Root;

dp[c][o->Id][s | o->n] += dp[p][q[r]->Id][s];

}

}

}

ll ans = 0;

for(int i = 0; i < t; i++)

ans += dp[c][q[i]->Id][S];

printf("%lld\n", ans);

if(ans > 42) return;

memset(F, 0, sizeof F);

for(int i = 0; i < N; i++)

F[V[0] = i] = true, DFS(1, lth[i]), F[i] = false;

for(int i = 0; i < ans; i++) 

res[R[i] = i][L] = '\0';

sort(R, R + ans, Cmp);

for(int i = 0; i < ans; i++)

printf("%s\n", res[R[i]]);

}

int main() {

Init();

Build();

Work();

return 0;

}

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1559: [JSOI2009]密码

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 675  Solved: 213
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Description

Input

Output

Sample Input

10 2
hello
world

Sample Output

2
helloworld
worldhello

HINT

Source

JSOI2009Day1