Gym - 101174E[AC自动机+DP]

版权声明：欢迎随便转载。 https://blog.csdn.net/a1214034447/article/details/89963623

题目链接:https://vjudge.net/problem/Gym-101174E

解题思路:

数据小，首先想可以不可以DP(虽然我一开始傻了没有往这个方向想)

将n个串建一个AC自动，然后用dp[i][j][k]表示选取了长度为i的密码长度，这个密码后缀对应自动机上的节点j，k是三个字符的选择情况的二进制，然后就是标记不可包含的节点，之后就可以在AC自动机上跑DP了。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mx = 1e4 + 10;
const int mod = 1000003;
int l,r,n,siz,rt,dp[30][mx][8];
int nxt[mx][26],fail[mx];
char str[100];
bool vis[mx];
int newnode(){
	memset(nxt[siz],0,sizeof(nxt[siz]));
	return siz++;
}
void insert(char *p){
	int now = rt;
	for(int i=0;p[i];i++){
		int id = p[i] - 'a';
		if(!nxt[now][id]) 
		nxt[now][id] = newnode();
		now = nxt[now][id];
	}
	vis[now] = 1;
}
void Getfail(){
	queue <int> q;
	for(int i=0;i<26;i++){
		if(!nxt[rt][i]) nxt[rt][i] = rt;
		else{
			fail[nxt[rt][i]] = rt;
			q.push(nxt[rt][i]);
		}
	}
	while(!q.empty()){
		int now = q.front();
		q.pop();
		vis[now] |= vis[fail[now]];
		for(int i=0;i<26;i++){
			if(!nxt[now][i]) nxt[now][i] = nxt[fail[now]][i];
			else{
				fail[nxt[now][i]] = nxt[fail[now]][i];
				q.push(nxt[now][i]);
			}
		} 
	}
}
int main(){
	rt = newnode();
	scanf("%d%d",&l,&r);
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%s",str);
		insert(str);
	}
	Getfail();
	dp[0][rt][0] = 1;
	for(int i=0;i<r;i++){
		for(int j=0;j<siz;j++){
			if(!vis[j])
			for(int k=0;k<8;k++){
				for(int w=0;w<26;w++){
					int fa = nxt[j][w];
					dp[i+1][fa][k|1] = (dp[i+1][fa][k|1]+dp[i][j][k])%mod;
					dp[i+1][fa][k|2] = (dp[i+1][fa][k|2]+dp[i][j][k])%mod;
				}
				int fa = nxt[j]['o'-'a'];
				dp[i+1][fa][k|4] = (dp[i+1][fa][k|4]+dp[i][j][k])%mod;
				fa = nxt[j]['i'-'a'];
				dp[i+1][fa][k|4] = (dp[i+1][fa][k|4]+dp[i][j][k])%mod;
				fa = nxt[j]['e'-'a'];
				dp[i+1][fa][k|4] = (dp[i+1][fa][k|4]+dp[i][j][k])%mod;
				fa = nxt[j]['s'-'a'];
				dp[i+1][fa][k|4] = (dp[i+1][fa][k|4]+dp[i][j][k])%mod;
				fa = nxt[j]['t'-'a'];
				dp[i+1][fa][k|4] = (dp[i+1][fa][k|4]+dp[i][j][k])%mod;
				dp[i+1][0][k|4] = (dp[i+1][0][k|4]+5*dp[i][j][k])%mod;
			}
		}
	}
	int ans = 0;
	for(int i=l;i<=r;i++){
		for(int j=0;j<siz;j++)
		if(!vis[j]) ans = (ans+dp[i][j][7])%mod;
	}
	printf("%d\n",ans);
    return 0;
}