CodeForces 86C-Genetic engineering

Genetic engineering

“Multidimensional spaces are completely out of style these days, unlike genetics problems” — thought physicist Woll and changed his subject of study to bioinformatics. Analysing results of sequencing he faced the following problem concerning DNA sequences. We will further think of a DNA sequence as an arbitrary string of uppercase letters “A”, “C”, “G” and “T” (of course, this is a simplified interpretation).

Let w be a long DNA sequence and s1, s2, …, sm — collection of short DNA sequences. Let us say that the collection filters w iff w can be covered with the sequences from the collection. Certainly, substrings corresponding to the different positions of the string may intersect or even cover each other. More formally: denote by |w| the length of w, let symbols of w be numbered from 1 to |w|. Then for each position i in w there exist pair of indices l, r (1 ≤ l ≤ i ≤ r ≤ |w|) such that the substring w[l … r] equals one of the elements s1, s2, …, sm of the collection.

Woll wants to calculate the number of DNA sequences of a given length filtered by a given collection, but he doesn’t know how to deal with it. Help him! Your task is to find the number of different DNA sequences of length n filtered by the collection {si}.

Answer may appear very large, so output it modulo 1000000009.

Input
First line contains two integer numbers n and m (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10) — the length of the string and the number of sequences in the collection correspondently.

Next m lines contain the collection sequences si, one per line. Each si is a nonempty string of length not greater than 10. All the strings consist of uppercase letters “A”, “C”, “G”, “T”. The collection may contain identical strings.

Output
Output should contain a single integer — the number of strings filtered by the collection modulo 1000000009 (109 + 9).

Example
Input
2 1
A
Output
1
Input
6 2
CAT
TACT
Output
2
Note
In the first sample, a string has to be filtered by “A”. Clearly, there is only one such string: “AA”.

In the second sample, there exist exactly two different strings satisfying the condition (see the pictures below).

参考书籍：《挑战程序设计竞赛》
**题目大意：**给定m个子串，求构造长n的母串的方案数。母串中每个字符都至少来自一个子串。
解题思路： AC自动机+动态规划
建立AC自动机，处理出每个节点的最长后缀模式串长度max_match.
定义状态dp[i][j][k]dp[i][j][k]dp[i][j][k]为构造长为i的母串，trie树节点为j，后缀从后往前数最远第k个字符不满足要求。
转移方程：设当前节点为cur，接受字符c转移到下一节点nxt
（1）若nxt->max_match>k
dp[length+1][nxt][0]+=dp[length][cur][k]
（2）否则，未匹配的尾部增长为k+1，dp[length+1][nxt][k+1]+=dp[length][cur][k]

#include<iostream>
#include<cstdio>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
using namespace std;
const int MAXN = 1e3+5;
const int MAXM = 15;
int dp[MAXN][MAXN][MAXM];
// dp[i][j][k] := 构造长i的母串，trie树节点为j，后缀有k个字符不满足要求
bool visited[MAXN][MAXN][MAXM];
const int MOD = 1e9+9 ;struct Node
{map<char, Node *> next;Node *fail;vector<int> match;static int _node_size;int id;int max_match;// 最长后缀匹配Node() : fail(NULL){id = _node_size++;max_match = 0;}
};int Node::_node_size = 0;Node *build(vector<string> pattens)
{Node *root = new Node();root->fail = root;for (int i = 0; i < pattens.size(); i++){Node *p = root;for (auto c : pattens[i]){if (p->next[c] == 0)p->next[c] = new Node();p = p->next[c];}p->match.push_back(i);p->max_match = pattens[i].size();}queue<Node *> que;for (int i = 0; i < 128; i++){if (!root->next[i]){root->next[i] = root;}else{root->next[i]->fail = root;que.push(root->next[i]);}}while (!que.empty()){Node *p = que.front();que.pop();for (auto a : p->next){int i = a.first;Node *np = a.second;if (!np)  continue;que.push(np);Node *f = p->fail;while (!f->next[i])f = f->fail;np->fail = f->next[i];np->max_match = max(np->max_match, np->fail->max_match);np->match.insert(np->match.end(), np->fail->match.begin(), np->fail->match.end());}}return root;
}Node *next_node(Node *p, char c)
{while (!p->next[c])p = p->fail;return p->next[c];
}Node *next_node(Node *p, string query)
{for(char c : query){p = next_node(p, c);}return p;
}struct State
{int length, k;Node *node;State(int length, Node *node, int k) : length(length), node(node), k(k) {}
};int main()
{int ids[256];char cs[4];const string acgt = "ACGT";for (int i = 0; i < acgt.length(); ++i){ids[acgt[i]] = i;cs[i] = acgt[i];}int n, m;cin>>n>>m;vector<string> dna(m);for (int i = 0; i < m; ++i){cin >> dna[i];}auto root = build(dna);dp[0][0][0] = 1;visited[0][0][0] = 1;queue<State> que;que.push(State(0, root, 0));while (!que.empty()){State s = que.front();que.pop();int length = s.length, k = s.k;Node *cur = s.node;if (length == n) continue;for (auto c : cs){auto nxt = next_node(cur, c);// 匹配到后缀且长度大于等于k+1，于是可以把k+1替换为0if (nxt->max_match > k){dp[length + 1][nxt->id][0] += dp[length][cur->id][k];dp[length + 1][nxt->id][0] %= MOD;if (!visited[length + 1][nxt->id][0]){visited[length + 1][nxt->id][0] = true;que.push(State(length + 1, nxt, 0));}}// 否则不行，未匹配的尾部增长为k+1else{if (k >= 9)continue;dp[length + 1][nxt->id][k + 1] += dp[length][cur->id][k];dp[length + 1][nxt->id][k + 1] %= MOD;if (!visited[length + 1][nxt->id][k + 1]){visited[length + 1][nxt->id][k + 1] = true;que.push(State(length + 1, nxt, k + 1));}}}}int ans = 0;for (int i = 0; i < root->_node_size; ++i){ans += dp[n][i][0];ans %= MOD;}cout<<ans<<endl;return 0;
}

CodeForces 86C-Genetic engineering相关推荐

Codeforces 86C Genetic engineering (AC自己主动机+dp)
题目大意: 要求构造一个串,使得这个串是由所给的串相连接构成,连接能够有重叠的部分. 思路分析: 首先用所给的串建立自己主动机,每一个单词节点记录当前节点可以达到的最长后缀. 開始的时候想的是dp[i ...
《挑战程序设计竞赛（第2版）》习题册攻略
本项目来源于GitHub 链接: 项目GitHub链接 1 前言项目为<挑战程序设计竞赛(第2版)>习题册攻略,已完结.可配合书籍或笔记,系统学习算法. 题量:约200道,代码注释内含详 ...
面对万亿级测序市场，纳米孔测序技术何去何从？
这是<肠道产业>第 482 篇文章 [直播预告]"Protein& Cell人类微生物组专刊线上论坛" 12月21日晚7点开播,敬请期待!(点击查看详情) 编者 ...
液态大脑与固态大脑——圣塔菲最新群体智能文集
来源:The Royal society 撰文 | Ricard Solé, Melanie Moses, and Stephanie Forrest 大脑,神经元构成的器官根植于许多生物体内.这是一 ...
XIV Open Cup named after E.V. Pankratiev. GP of Europe
A. The Motorway 等价于找到最小和最大的$L$满足存在$S$使得$S+(i-1)L\leq a_i\leq S+i\times L$ 即 $S\leq\min((1-i)L+a_i)$ ...
备战ccpc分站赛：秦皇岛和威海站（数论模块和dp模块）
挑战程序设计竞赛(第2版)练习题 tips:难度(个人主观判断): 简单* 简单但卡思维 ** 中 *** 中稍加思考 **** 难 ***** 1 . 记录结果再利用的"动态规划" ...
[渝粤教育] 南开大学思辨式英文写作参考资料
教育 -思辨式英文写作-章节资料考试资料-南开大学[] 随堂小测:What are the characteristics of critical essays? 1.[多选题]Which of th ...
【渝粤题库】陕西师范大学200481 高级英语（一）
<高级英语(一)>作业 I. Explain the italicized words in English 1.The very act -was for me a far greate ...
matlab 合成生物学,合成生物学原理
你将学到什么 Modern techniques in DNA assembly and regulation of gene expression and protein activity How ...

CodeForces 86C-Genetic engineering

Genetic engineering

CodeForces 86C-Genetic engineering相关推荐

最新文章

热门文章