POJ 1625 Censored!(AC自动机+DP+高精度)
Time Limit: 5000MS | Memory Limit: 10000K | |
Total Submissions: 6956 | Accepted: 1887 |
Description
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 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
Sample Input
2 3 1
ab
bb
Sample Output
5
Source
//============================================================================
// Name : POJ.cpp
// Author :
// Version :
// Copyright : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================ #include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <queue>
#include <map>
using namespace std;
map<char,int>mp;
int N,M,P;
struct Matrix
{
int mat[][];
int n;
Matrix(){}
Matrix(int _n)
{
n=_n;
for(int i = ;i < n;i++)
for(int j = ;j < n;j++)
mat[i][j] = ;
}
};
struct Trie
{
int next[][],fail[];
bool end[];
int L,root;
int newnode()
{
for(int i = ;i < ;i++)
next[L][i] = -;
end[L++] = false;
return L-;
}
void init()
{
L = ;
root = newnode();
}
void insert(char buf[])
{
int len = strlen(buf);
int now = root;
for(int i = ;i < len;i++)
{
if(next[now][mp[buf[i]]] == -)
next[now][mp[buf[i]]] = newnode();
now = next[now][mp[buf[i]]];
}
end[now] = true;
}
void build()
{
queue<int>Q;
fail[root] = root;
for(int i = ;i < ;i++)
if(next[root][i] == -)
next[root][i] = root;
else
{
fail[next[root][i]] = root;
Q.push(next[root][i]);
}
while(!Q.empty())
{
int now = Q.front();
Q.pop();
if(end[fail[now]]==true)end[now]=true;
for(int i = ;i < ;i++)
if(next[now][i] == -)
next[now][i] = next[fail[now]][i];
else
{
fail[next[now][i]] = next[fail[now]][i];
Q.push(next[now][i]);
}
}
}
Matrix getMatrix()
{
Matrix res = Matrix(L);
for(int i = ;i < L;i++)
for(int j = ;j < N;j++)
if(end[next[i][j]]==false)
res.mat[i][next[i][j]]++;
return res;
}
void debug()
{
for(int i = ;i < L;i++)
{
printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
for(int j = ;j < ;j++)
printf("%2d",next[i][j]);
printf("]\n");
}
} }; /*
* 高精度,支持乘法和加法
*/
struct BigInt
{
const static int mod = ;
const static int DLEN = ;
int a[],len;
BigInt()
{
memset(a,,sizeof(a));
len = ;
}
BigInt(int v)
{
memset(a,,sizeof(a));
len = ;
do
{
a[len++] = v%mod;
v /= mod;
}while(v);
}
BigInt(const char s[])
{
memset(a,,sizeof(a));
int L = strlen(s);
len = L/DLEN;
if(L%DLEN)len++;
int index = ;
for(int i = L-;i >= ;i -= DLEN)
{
int t = ;
int k = i - DLEN + ;
if(k < )k = ;
for(int j = k;j <= i;j++)
t = t* + s[j] - '';
a[index++] = t;
}
}
BigInt operator +(const BigInt &b)const
{
BigInt res;
res.len = max(len,b.len);
for(int i = ;i <= res.len;i++)
res.a[i] = ;
for(int i = ;i < res.len;i++)
{
res.a[i] += ((i < len)?a[i]:)+((i < b.len)?b.a[i]:);
res.a[i+] += res.a[i]/mod;
res.a[i] %= mod;
}
if(res.a[res.len] > )res.len++;
return res;
}
BigInt operator *(const BigInt &b)const
{
BigInt res;
for(int i = ; i < len;i++)
{
int up = ;
for(int j = ;j < b.len;j++)
{
int temp = a[i]*b.a[j] + res.a[i+j] + up;
res.a[i+j] = temp%mod;
up = temp/mod;
}
if(up != )
res.a[i + b.len] = up;
}
res.len = len + b.len;
while(res.a[res.len - ] == &&res.len > )res.len--;
return res;
}
void output()
{
printf("%d",a[len-]);
for(int i = len-;i >= ;i--)
printf("%04d",a[i]);
printf("\n");
}
};
char buf[];
BigInt dp[][];
Trie ac;
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout); while(scanf("%d%d%d",&N,&M,&P)==)
{
gets(buf);
gets(buf);
mp.clear();
int len = strlen(buf);
for(int i = ;i < len;i++)
mp[buf[i]]=i;
ac.init();
for(int i = ;i < P;i++)
{
gets(buf);
ac.insert(buf);
}
ac.build();
// ac.debug();
Matrix a= ac.getMatrix(); // for(int i = 0;i <a.n;i++)
// {
// for(int j=0;j<a.n;j++)printf("%d ",a.mat[i][j]);
// cout<<endl;
// } int now = ;
dp[now][] = ;
for(int i = ;i < a.n;i++)
dp[now][i] = ;
for(int i = ;i < M;i++)
{
now^=;
for(int j = ;j < a.n;j++)
dp[now][j] = ;
for(int j = ;j < a.n;j++)
for(int k = ;k < a.n;k++)
if(a.mat[j][k] > )
dp[now][k] = dp[now][k]+dp[now^][j]*a.mat[j][k];
// for(int j = 0;j < a.n;j++)
// dp[now][j].output();
}
BigInt ans = ;
for(int i = ;i < a.n;i++)
ans = ans + dp[now][i];
ans.output();
}
return ;
}
POJ 1625 Censored!(AC自动机+DP+高精度)的更多相关文章
- Match:Censored!(AC自动机+DP+高精度)(POJ 1625)
Censored! 题目大意:给定一些字符,将这些字符组成一个固定长度的字符串,但是字符串不能包含一些禁词,问你有多少种组合方式. 这是一道好题,既然出现了“一些”禁词,那么这题肯定和AC自动机有点 ...
- POJ 1625 Censored! [AC自动机 高精度]
Censored! Time Limit: 5000MS Memory Limit: 10000K Total Submissions: 9793 Accepted: 2686 Descrip ...
- POJ 1625 Censored ( Trie图 && DP && 高精度 )
题意 : 给出 n 个单词组成的字符集 以及 p 个非法串,问你用字符集里面的单词构造长度为 m 的单词的方案数有多少种? 分析 :先构造出 Trie 图方便进行状态转移,这与在 POJ 2278 中 ...
- [POJ1625]Censored!(AC自动机+DP+高精度)
Censored! Time Limit: 5000MS Memory Limit: 10000K Total Submissions: 10824 Accepted: 2966 Descri ...
- poj 1625 (AC自动机好模版,大数好模版)
题目 给n个字母,构成长度为m的串,总共有n^m种.给p个字符串,问n^m种字符串中不包含(不是子串)这p个字符串的个数. 将p个不能包含的字符串建立AC自动机,每个结点用val值来标记以当前节点为后 ...
- POJ1625 Censored! —— AC自动机 + DP + 大数
题目链接:https://vjudge.net/problem/POJ-1625 Censored! Time Limit: 5000MS Memory Limit: 10000K Total S ...
- POJ1625 Censored!(AC自动机+DP)
题目问长度m不包含一些不文明单词的字符串有多少个. 依然是水水的AC自动机+DP..做完后发现居然和POJ2778是一道题,回过头来看都水水的... dp[i][j]表示长度i(在自动机转移i步)且后 ...
- 对AC自动机+DP题的一些汇总与一丝总结 (2)
POJ 2778 DNA Sequence (1)题意 : 给出m个病毒串,问你由ATGC构成的长度为 n 且不包含这些病毒串的个数有多少个 关键字眼:不包含,个数,长度 DP[i][j] : 表示长 ...
- HDU 2457 DNA repair(AC自动机+DP)题解
题意:给你几个模式串,问你主串最少改几个字符能够使主串不包含模式串 思路:从昨天中午开始研究,研究到现在终于看懂了.既然是多模匹配,我们是要用到AC自动机的.我们把主串放到AC自动机上跑,并保证不出现 ...
随机推荐
- ios 关于使用异步网络请求时block回调的内存注意
在一个controller中,使用 NSURLSessionDataTask *dataTask = [[NSURLSession sharedSession] dataTaskWithRequest ...
- pc/app 项目/功能设计
2015-08-11 15:29:59 首先, 提供服务的整个系统包含哪几个设备 用户->[PC浏览器|APP|APP浏览器]->web服务器->[PHP/JAVA组件]->[ ...
- c++ template
在类中其中一个函数使用模板,函数定义和实现必须放在头文件里. ca.h template<typename T> void swap2(T &a,T &b) { T c=a ...
- jquery 常用的备忘
1.table 取 tr 的 id值 <table class="table table-striped table-hover table-bordered" id=&q ...
- Effective C++ -----条款31:将文件间的编译依存关系降至最低
支持“编译依存性最小化”的一般构想是:相依于声明式,不要相依于定义式.基于此构想的两个手段是Handle classes 和 Interface classes. 程序库头文件应该以“完全且仅有声明式 ...
- 【leetcode】Word Search II(hard)★
Given a 2D board and a list of words from the dictionary, find all words in the board. Each word mus ...
- Java发送邮件初窥
一.背景 最近朋友的公司有用到这个功能,之前对这一块也不是很熟悉,就和他一起解决出现的异常的同时,也初窥一下使用Apache Common Email组件进行邮件发送. 二.Java发送邮件的注意事项 ...
- Maven Java EE Configuration Problem 的完美解决办法
背景: 最近在修改项目的时候,发现修改了项目依赖以后会出现如下图:Maven Java EE Configuration Problem 的问题,对于有强迫症的我来说,看到项目上面有个很小的红色小叉号 ...
- device host global 函数要求
转自:https://kheresy.wordpress.com/2007/11/05/nvidia-cuda-api%EF%BC%88%E4%B8%8A%EF%BC%89/ Function typ ...
- javaWeb四大域对象
servletContext Request HttpSession pageContext 详细参考:http://www.tuicool.com/articles/NJfyMrn