題意：
給定你n 種字符， 和p 個禁止串， 問你長度爲m 的串有多少種 不包含 禁止串？
思路：
AC自動機是顯然的。
矩陣快速冪也可以 ，dp 也可以。
但是這是個大數。 極限答案有85位左右。
我用快速冪寫了一發，不是爆內存就是因爲內存開的過大 小樣例都出不來。
只能考慮dp了。
dp 狀態很好想:
令dp[i][j] 表示目前走 第i 位長度， 在自動機上j 節點的方案數。
轉移就是上一個節點 轉下一個節點了。
dp[i][j] += dp[i-1][k]
爲了節省空間 搞成滾動數組即可
不過看了看討論區， 說是 字符串有負數出現 就是 輸入 的ASCII碼會在128~255之間？ 但是我的並沒有考慮這種情況， 也A了= =（maybe數據改掉了？）
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <string>
#include <istream>
#include <ostream>
using namespace std;
///高精度
const int MAXN = 90;
struct bign
{
int len, s[MAXN];
bign ()
{
memset(s, 0, sizeof(s));
len = 1;
}
bign (int num) { *this = num; }
bign (const char *num) { *this = num; }
bign operator = (const int num)
{
char s[MAXN];
sprintf(s, "%d", num);
*this = s;
return *this;
}
bign operator = (const char *num)
{
for(int i = 0; num[i] == '0'; num++) ; //去前導0
len = strlen(num);
for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
return *this;
}
bign operator + (const bign &b) const //+
{
bign c;
c.len = 0;
for(int i = 0, g = 0; g || i < max(len, b.len); i++)
{
int x = g;
if(i < len) x += s[i];
if(i < b.len) x += b.s[i];
c.s[c.len++] = x % 10;
g = x / 10;
}
return c;
}
bign operator += (const bign &b)
{
*this = *this + b;
return *this;
}
void clean()
{
while(len > 1 && !s[len-1]) len--;
}
string str() const
{
string res = "";
for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;
return res;
}
};
istream& operator >> (istream &in, bign &x)
{
string s;
in >> s;
x = s.c_str();
return in;
}
ostream& operator << (ostream &out, const bign &x)
{
out << x.str();
return out;
}
///
const int maxn = 100 + 3;
char s[100];
int mp[128 ];
int fmp[128];
bign dp[2][maxn];
const int inf = 0x3f3f3f3f;
struct Trie{
int L, root;
int next[maxn][50];
int fail[maxn];
int flag[maxn];
int Hash[maxn];
int fHash[maxn];
int cnt;
int mx;
void init(int mx_){
L = 0;
mx = mx_;
root = newnode();
cnt = 0;
}
int newnode(){
for (int i = 0; i < mx; ++i){
next[L][i] = -1;
}
flag[L] = 0;
return L++;
}
void insert(char* s){
int len = strlen(s);
int nod = root;
for (int i = 0; i < len; ++i){
int id = fmp[s[i] ];
if (next[nod][id] == -1){
next[nod][id] = newnode();
}
nod = next[nod][id];
}
flag[nod] = 1;
}
void bfs(){
fail[root] = root;
queue<int>q;
for (int i = 0; i < mx; ++i){
if (next[root][i] == -1){
next[root][i] = root;
}
else {
fail[next[root][i] ] = root;
q.push(next[root][i]);
}
}
while(!q.empty()){
int u = q.front(); q.pop();
for (int i = 0; i < mx; ++i){
if (next[u][i] == -1){
next[u][i] = next[fail[u] ][i];
}
else {
fail[next[u][i] ] = next[fail[u] ][i];
q.push(next[u][i]);
}
}
}
}
void deal(){
for (int i = 0; i < L; ++i){
if (flag[i]) continue;
int tmp = i;
while(tmp != root){
if (flag[tmp]){
flag[i] = 1;
break;
}
tmp = fail[tmp];
}
if (!flag[i]){
Hash[cnt++] = i;
fHash[i] = cnt - 1;
}
}
}
void solve(int n){
for (int j = 0; j < maxn; ++j){
dp[0][j] = 0;
}
dp[0][0] = 1;
int foo = 0;
for (int k = 1; k <= n; ++k){
int first = foo;
int second = foo ^ 1;
for (int i = 0; i < maxn; ++i){
dp[second][i] = 0;
}
for (int i = 0; i < cnt; ++i){
for (int j = 0; j < mx; ++j){
int now = Hash[i];
int nx = next[now][j];
if (flag[nx]) continue;
dp[second][nx] += dp[first][now];
}
}
foo ^= 1;
}
bign ans = 0;
for (int i = 0; i < cnt; ++i){
int nod = Hash[i];
ans += dp[foo][nod];
}
string tt = ans.str();
if (tt.length() == 0)cout<<0<<endl;
else cout << ans << endl;
}
}ac;
/**
8881784197001252323389053344726562500000000000000000000000000000000000000000000000000
**/
int main(){
int n, m, p;
scanf("%d %d %d",&n, &m, &p);
scanf("%s", s);
int len = strlen(s);
for (int i = 0; i < len; ++i){
mp[i] = s[i];
fmp[s[i]] = i;
}
ac.init(len);
for (int i = 0; i < p; ++i){
scanf("%s", s);
ac.insert(s);
}
ac.bfs();
ac.deal();
ac.solve(m);
return 0;
}
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Censored!
Description
The alphabet of Freeland consists of exactly N letters. Each sentence of Freeland language (also known as Freish) consists of exactly M letters without word breaks. So, there exist exactly N^M different Freish sentences.
But after recent election of Mr. Grass Jr. as Freeland president some words offending him were declared unprintable and all sentences containing at least one of them were forbidden. The sentence S contains a word W if W is a substring of S i.e. exists such k >= 1 that S[k] = W[1], S[k+1] = W[2], ...,S[k+len(W)-1] = W[len(W)], where k+len(W)-1 <= M and len(W) denotes length of W. Everyone who uses a forbidden sentence is to be put to jail for 10 years. Find out how many different sentences can be used now by freelanders without risk to be put to jail for using it. Input
The first line of the input file contains three integer numbers: N -- the number of letters in Freish alphabet, M -- the length of all Freish sentences and P -- the number of forbidden words (1 <= N <= 50, 1 <= M <= 50, 0 <= P <= 10).
The second line contains exactly N different characters -- the letters of the Freish alphabet (all with ASCII code greater than 32). The following P lines contain forbidden words, each not longer than min(M, 10) characters, all containing only letters of Freish alphabet. Output
Output the only integer number -- the number of different sentences freelanders can safely use.
Sample Input 2 3 1 ab bb Sample Output 5 Source
Northeastern Europe 2001, Northern Subregion
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