Ural1158 看上去很困难的一道题。
原文地址 http://blog.****.net/prolightsfxjh/article/details/54729646
题意：给出n个不同的字符，用这n个字符构成长度为m的字符串，要求每个串的子串都不出现给定的p个串中的任一个，求满足要求的字符串的个数。
AC自动机+dp
因为构成的最终串是由一个字符一个字符添加到字符串尾部构成的，那么如果一个串的后缀如果恰好是某个给定串的前缀时，这个串就可能最终成为非法串。
用k个给定串建立AC自动机，然后从根节点开始递推，
dpij表示递推到第j个字符当前在自动机上的i号节点时的方案数，如果下一个节点是k，且不是危险节点，则把dpij加到dp[k][i+1]里，
跑一遍，然后答案就是所有非危险节点的方案数的和(其实危险节点上都是0)。因为危险节点是给定串的终点或者其后缀节点是危险节点的点，遍历到危险节点的点上的方案必定是包含了给定串的方案，故不能记录这些。
此外这里dpij会很多，故要用高精度整数。笔者自己收藏的大整数类的版，可行长太长了，所以MLE了一发，调小了才过。⊙﹏⊙‖∣
复杂度 O(n^3)
模版串长度很小，所以Trie上节点很少，递推成本也就比较低。
其实这道题就是三个部分， 实现了AC自动机和大整数， 就是一道DP题了，分开来看 理解了KMP算法和Trie树 AC自动机很简单， 大整数朴素的O（n^2）实现也很简单，DP方程也很简单，但是组合起来，就很困难了。
诶 东北欧赛区的题，总是这么~
代码如下 洋洋洒洒12k 当然不是我写的， 不过这份代码非常清晰易懂，反正智障的我看的挺明白。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <map>
using namespace std;
const int CHAR_SIZE = 51;
const int MAX_SIZE = 105;
map<char, int> mp; struct AC_Machine{
int ch[MAX_SIZE][CHAR_SIZE], danger[MAX_SIZE], fail[MAX_SIZE];
int sz; void init(){
sz = 1;
memset(ch[0], 0, sizeof ch[0]);
memset(danger, 0, sizeof danger);
} void _insert(char *s){
int n = strlen(s);
int u = 0, c;
for(int i = 0; i < n; i++){
c = mp[s[i]];
if(!ch[u][c]){
memset(ch[sz], 0, sizeof ch[sz]);
danger[sz] = 0;
ch[u][c] = sz++;
}
u = ch[u][c];
}
danger[u] = 1;
} void _build(){
queue<int> Q;
fail[0] = 0;
for(int c = 0, u; c < CHAR_SIZE; c++){
u = ch[0][c];
if(u){Q.push(u); fail[u] = 0;}
}
int r;
while(!Q.empty()){
r = Q.front();
Q.pop();
danger[r] |= danger[fail[r]];
for(int c = 0, u; c < CHAR_SIZE; c++){
u = ch[r][c];
if(!u){ch[r][c] = ch[fail[r]][c]; continue; }
fail[u] = ch[fail[r]][c];
Q.push(u);
}
}
}
}ac; char s[MAX_SIZE]; #include <string>
#include <iostream>
#include <iosfwd>
#include <cmath>
#include <cstring>
#include <stdlib.h>
#include <stdio.h>
#include <cstring>
#define MAX_L 205 //最大长度，可以修改
using namespace std; class bign
{
public:
int len, s[MAX_L];//数的长度，记录数组
//构造函数
bign();
bign(const char*);
bign(int);
bool sign;//符号 1正数 0负数
string toStr() const;//转化为字符串，主要是便于输出
friend istream& operator>>(istream &,bign &);//重载输入流
friend ostream& operator<<(ostream &,bign &);//重载输出流
//重载复制
bign operator=(const char*);
bign operator=(int);
bign operator=(const string);
//重载各种比较
bool operator>(const bign &) const;
bool operator>=(const bign &) const;
bool operator<(const bign &) const;
bool operator<=(const bign &) const;
bool operator==(const bign &) const;
bool operator!=(const bign &) const;
//重载四则运算
bign operator+(const bign &) const;
bign operator++();
bign operator++(int);
bign operator+=(const bign&);
bign operator-(const bign &) const;
bign operator--();
bign operator--(int);
bign operator-=(const bign&);
bign operator*(const bign &)const;
bign operator*(const int num)const;
bign operator*=(const bign&);
bign operator/(const bign&)const;
bign operator/=(const bign&);
//四则运算的衍生运算
bign operator%(const bign&)const;//取模（余数）
bign factorial()const;//阶乘
bign Sqrt()const;//整数开根（向下取整）
bign pow(const bign&)const;//次方
//一些乱乱的函数
void clean();
~bign();
};
#define max(a,b) a>b ? a : b
#define min(a,b) a<b ? a : b bign::bign()
{
memset(s, 0, sizeof(s));
len = 1;
sign = 1;
} bign::bign(const char *num)
{
*this = num;
} bign::bign(int num)
{
*this = num;
} string bign::toStr() const
{
string res;
res = "";
for (int i = 0; i < len; i++)
res = (char)(s[i] + '0') + res;
if (res == "")
res = "0";
if (!sign&&res != "0")
res = "-" + res;
return res;
} istream &operator>>(istream &in, bign &num)
{
string str;
in>>str;
num=str;
return in;
} ostream &operator<<(ostream &out, bign &num)
{
out<<num.toStr();
return out;
} bign bign::operator=(const char *num)
{
memset(s, 0, sizeof(s));
char a[MAX_L] = "";
if (num[0] != '-')
strcpy(a, num);
else
for (int i = 1; i < strlen(num); i++)
a[i - 1] = num[i];
sign = !(num[0] == '-');
len = strlen(a);
for (int i = 0; i < strlen(a); i++)
s[i] = a[len - i - 1] - 48;
return *this;
} bign bign::operator=(int num)
{
char temp[MAX_L];
sprintf(temp, "%d", num);
*this = temp;
return *this;
} bign bign::operator=(const string num)
{
const char *tmp;
tmp = num.c_str();
*this = tmp;
return *this;
} bool bign::operator<(const bign &num) const
{
if (sign^num.sign)
return num.sign;
if (len != num.len)
return len < num.len;
for (int i = len - 1; i >= 0; i--)
if (s[i] != num.s[i])
return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i]));
return !sign;
} bool bign::operator>(const bign&num)const
{
return num < *this;
} bool bign::operator<=(const bign&num)const
{
return !(*this>num);
} bool bign::operator>=(const bign&num)const
{
return !(*this<num);
} bool bign::operator!=(const bign&num)const
{
return *this > num || *this < num;
} bool bign::operator==(const bign&num)const
{
return !(num != *this);
} bign bign::operator+(const bign &num) const
{
if (sign^num.sign)
{
bign tmp = sign ? num : *this;
tmp.sign = 1;
return sign ? *this - tmp : num - tmp;
}
bign result;
result.len = 0;
int temp = 0;
for (int i = 0; temp || i < (max(len, num.len)); i++)
{
int t = s[i] + num.s[i] + temp;
result.s[result.len++] = t % 10;
temp = t / 10;
}
result.sign = sign;
return result;
} bign bign::operator++()
{
*this = *this + 1;
return *this;
} bign bign::operator++(int)
{
bign old = *this;
++(*this);
return old;
} bign bign::operator+=(const bign &num)
{
*this = *this + num;
return *this;
} bign bign::operator-(const bign &num) const
{
bign b=num,a=*this;
if (!num.sign && !sign)
{
b.sign=1;
a.sign=1;
return b-a;
}
if (!b.sign)
{
b.sign=1;
return a+b;
}
if (!a.sign)
{
a.sign=1;
b=bign(0)-(a+b);
return b;
}
if (a<b)
{
bign c=(b-a);
c.sign=false;
return c;
}
bign result;
result.len = 0;
for (int i = 0, g = 0; i < a.len; i++)
{
int x = a.s[i] - g;
if (i < b.len) x -= b.s[i];
if (x >= 0) g = 0;
else
{
g = 1;
x += 10;
}
result.s[result.len++] = x;
}
result.clean();
return result;
} bign bign::operator * (const bign &num)const
{
bign result;
result.len = len + num.len; for (int i = 0; i < len; i++)
for (int j = 0; j < num.len; j++)
result.s[i + j] += s[i] * num.s[j]; for (int i = 0; i < result.len; i++)
{
result.s[i + 1] += result.s[i] / 10;
result.s[i] %= 10;
}
result.clean();
result.sign = !(sign^num.sign);
return result;
} bign bign::operator*(const int num)const
{
bign x = num;
bign z = *this;
return x*z;
}
bign bign::operator*=(const bign&num)
{
*this = *this * num;
return *this;
} bign bign::operator /(const bign&num)const
{
bign ans;
ans.len = len - num.len + 1;
if (ans.len < 0)
{
ans.len = 1;
return ans;
} bign divisor = *this, divid = num;
divisor.sign = divid.sign = 1;
int k = ans.len - 1;
int j = len - 1;
while (k >= 0)
{
while (divisor.s[j] == 0) j--;
if (k > j) k = j;
char z[MAX_L];
memset(z, 0, sizeof(z));
for (int i = j; i >= k; i--)
z[j - i] = divisor.s[i] + '0';
bign dividend = z;
if (dividend < divid) { k--; continue; }
int key = 0;
while (divid*key <= dividend) key++;
key--;
ans.s[k] = key;
bign temp = divid*key;
for (int i = 0; i < k; i++)
temp = temp * 10;
divisor = divisor - temp;
k--;
}
ans.clean();
ans.sign = !(sign^num.sign);
return ans;
} bign bign::operator/=(const bign&num)
{
*this = *this / num;
return *this;
} bign bign::operator%(const bign& num)const
{
bign a = *this, b = num;
a.sign = b.sign = 1;
bign result, temp = a / b*b;
result = a - temp;
result.sign = sign;
return result;
} bign bign::pow(const bign& num)const
{
bign result = 1;
for (bign i = 0; i < num; i++)
result = result*(*this);
return result;
} bign bign::factorial()const
{
bign result = 1;
for (bign i = 1; i <= *this; i++)
result *= i;
return result;
} void bign::clean()
{
if (len == 0) len++;
while (len > 1 && s[len - 1] == '\0')
len--;
} bign bign::Sqrt()const
{
if(*this<0)return -1;
if(*this<=1)return *this;
bign l=0,r=*this,mid;
while(r-l>1)
{
mid=(l+r)/2;
if(mid*mid>*this)
r=mid;
else
l=mid;
}
return l;
} bign::~bign()
{
} bign dp[MAX_SIZE][CHAR_MAX]; int main()
{ freopen("t.txt", "r", stdin);
#ifdef LOCAL //freopen("1.out", "w", stdout);
int T = 1;
while(T--){
#endif // LOCAL
//ios::sync_with_stdio(false); cin.tie(0); int n, m, p;
scanf("%d%d%d", &n, &m, &p);
scanf("%s", s);
for(int i = 0; i < n; i++){mp[s[i]] = i;}
ac.init();
while(p--){
scanf("%s", s);
ac._insert(s);
}
int i, j, k;
for(i = 0; i < ac.sz; i++){
for(j = 0; j <= m; j++){
dp[i][j] = 0;
}
}
ac._build();
dp[0][0] = 1;
for(i = 1; i <= m; i++){
for(j = 0; j < ac.sz; j++){
for(k = 0; k < n; k++){
if(!ac.danger[ac.ch[j][k]]){
dp[ac.ch[j][k]][i] += dp[j][i-1];
}
}
}
}
bign ans = 0;
for(i = 0; i < ac.sz; i++){ans += dp[i][m];}
cout << ans << endl; #ifdef LOCAL
cout << endl;
}
#endif // LOCAL
return 0;
}