1625

AC自动机 + DP + 高精，设dp[i, j]表示处于句子的第i个字母，AC自动机的状态j，则dp[i, j] = sum(dp[i - 1, "a + b"])("a + b"表示在AC状态机中，状态j是从状态a通过字母b跳过来的，这里不可以编辑公式，我想不出好的表达方式，见谅)

这条题有一个trick，比如有两个forbidden words，asfgkl和fg，后面是前面的字串，则在状态转移时asfg其实就不要再往下转了

 

#include <iostream>#include <vector>#include <string>#include <iterator>#include <algorithm>using namespace std;class largenum{public:largenum(): num(1, 0){}vector<unsigned char> num;void operator = (const int x);void operator +=(const largenum &x);friend ostream &operator << (ostream &op, largenum &x);};void largenum::operator = (const int x){num.clear();num.push_back(x);}void largenum::operator +=(const largenum &x){unsigned char carry = 0;if(num.size() < x.num.size()){int i = 0;for(; i < num.size(); ++i){num[i] += x.num[i] + carry;if(num[i] >= 10){num[i] -= 10;carry = 1;}else{carry = 0;}}for(; i < x.num.size(); ++i){num.push_back(x.num[i] + carry);if(num[i] >= 10){num[i] -= 10;carry = 1;}else{carry = 0;}}if(carry){num.push_back(1);}}else{int i = 0;for(; i < x.num.size(); ++i){num[i] += x.num[i] + carry;if(num[i] >= 10){num[i] -= 10;carry = 1;}else{carry = 0;}}for(; i < num.size() && carry; ++i){++num[i];if(num[i] >= 10){num[i] -= 10;}else{carry = 0;}}if(carry){num.push_back(1);}}}ostream &operator << (ostream &op, largenum &x){string temp;for(vector<unsigned char>::reverse_iterator iter = x.num.rbegin(); iter != x.num.rend(); ++iter){temp.push_back(*iter + '0');}cout << temp;return op;}struct tnode{static const int CNUM = 50;tnode(): fail(NULL), isword(false){memset(child, NULL, sizeof(child));}tnode *child[CNUM];tnode *fail;bool isword;};int N, M, P, co = 1;unsigned char letters[51], censored[10][51], fb[10][50], len[10];tnode mem[501], &root = mem[0];largenum dp1[500], dp2[500], *current, *next, ans;void init(){cin >> N >> M >> P >> letters;for(int i = 0; i < P; cin >> censored[i], ++i);for(int i = 0, j; i < P; ++i){for(j = 0; censored[i][j]; ++j){fb[i][j] = find(letters, letters + N, censored[i][j]) - letters;}len[i] = j;}}void insert(){tnode *p;for(int i = 0; i < P; ++i){p = &root;for(int j = 0; j < len[i]; ++j){if(!p ->child[fb[i][j]]){p ->child[fb[i][j]] = &mem[co++];}p = p ->child[fb[i][j]];}p ->isword = true;}}void build_AC(){tnode *queue[510], *p;int head = 0, rear = 1;queue[0] = &root;root.fail = &root;while(head < rear){for(int i = 0; i < tnode::CNUM; ++i){if(queue[head] ->child[i]){if(&root == queue[head]){queue[head] ->child[i] ->fail = &root;}else{p = queue[head] ->fail;while(!p ->child[i] && p != &root){p = p ->fail;}queue[head] ->child[i] ->fail = p ->child[i]? p ->child[i]: &root;if(queue[head] ->child[i] ->fail ->isword || queue[head] ->isword){queue[head] ->child[i] ->isword = true;}}queue[rear] = queue[head] ->child[i];++rear;}}++head;}}int main(){init();insert();build_AC();dp1[0] = 1;current = dp1;next = dp2;for(int i = 0; i < M; ++i){for(int j = 0; j < co; ++j){for(int k = 0; k < N; ++k){if(!mem[j].isword){tnode *p = &mem[j];while(!p ->child[k] && &root != p){p = p ->fail;}p = p ->child[k]? p ->child[k]: &root;if(!p ->isword){next[p - mem] += current[j];}}}}swap(current, next);for(int i = 0; i < co; next[i] = 0, ++i);}for(int i = 0; i < co; ++i){ans += current[i];}cout << ans << endl;return 0;}