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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
[Submit][Status][Discuss]Description
Input
Output
Sample Input
10 2
hello
world
Sample Output
2
helloworld
worldhello
HINT
Source
BZOJ 1559: [JSOI2009]密码( AC自动机 + 状压dp )
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原文地址：http://www.cnblogs.com/JSZX11556/p/5153664.html
