问题:Biologists finally invent techniques of repairing DNA that contains segments causing kinds of inherited diseases. For the sake of simplicity, a DNA is represented as a string containing characters 'A', 'G' , 'C' and 'T'. The repairing techniques are simply to change some characters to eliminate all segments causing diseases. For example, we can repair a DNA "AAGCAG" to "AGGCAC" to eliminate the initial causing disease segments "AAG", "AGC" and "CAG" by changing two characters. Note that the repaired DNA can still contain only characters 'A', 'G', 'C' and 'T'.

You are to help the biologists to repair a DNA by changing least number of characters.
 
Input
The input consists of multiple test cases. Each test case starts with a line containing one integers N (1 ≤ N ≤ 50), which is the number of DNA segments causing inherited diseases.
The following N lines gives N non-empty strings of length not greater than 20 containing only characters in "AGCT", which are the DNA segments causing inherited disease.
The last line of the test case is a non-empty string of length not greater than 1000 containing only characters in "AGCT", which is the DNA to be repaired.

The last test case is followed by a line containing one zeros.
 
Output
For each test case, print a line containing the test case number( beginning with 1) followed by the
number of characters which need to be changed. If it's impossible to repair the given DNA, print -1.

Sample Input
2
AAA
AAG
AAAG    
2
A
TG
TGAATG
4
A
G
C
T
AGT
0

Sample Output
Case 1: 1
Case 2: 4
Case 3: -1

回答:题意给出一些不合法的模式DNA串,给出一个原串,问最少需要修改多少个字符,使得原串中不包含非法串
多串匹配,先想到AC自动机,需要求出最少需要修改多少字符,DP。
结合在一起
每一次沿着Trie树往下走,不能到达叶子结点罢了。不过对于为空但是合法的孩子需要进行处理。
DP方面,dp[i][j]表示前i个字符,当前为状态j的时候,需要修改的最少字符数。
从i-1的状态,找到之后的状态,如果字符与原串相同,则不变,否则+1。代码如下:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define N 100005
#define MOD 100000
#define inf 1<<29
#define LL long long
using namespace std;
struct Trie{
    Trie *next[4];
    Trie *fail;
    int kind,isword;
};
Trie *que[N],s[N];
int idx;
int id(char ch){
    if(ch=='A') return 0;
    else if(ch=='T') return 1;
    else if(ch=='C') return 2;
    return 3;
}
Trie *NewNode(){
    Trie *tmp=&s[idx];
    for(int i=0;i<4;i++) tmp->next[i]=NULL;
    tmp->isword=0;
    tmp->kind=idx++;
    tmp->fail=NULL;
    return tmp;
}
void Insert(Trie *root,char *s,int len){
    Trie *p=root;
    for(int i=0;i<len;i++){
        if(p->next[id(s[i])]==NULL)
            p->next[id(s[i])]=NewNode();
        p=p->next[id(s[i])];
    }
    p->isword=1;
}
void Bulid_Fail(Trie *root){
    int head=0,tail=0;
    que[tail++]=root;
    root->fail=NULL;
    while(head<tail){
        Trie *tmp=que[head++];
        for(int i=0;i<4;i++){
            if(tmp->next[i]){
                if(tmp==root) tmp->next[i]->fail=root;
                else{
                    Trie *p=tmp->fail;
                    while(p!=NULL){
                        if(p->next[i]){
                           tmp->next[i]->fail=p->next[i];
                           break;
                        }
                        p=p->fail;
                    }
                    if(p==NULL) tmp->next[i]->fail=root;
                }
                if(tmp->next[i]->fail->isword) tmp->next[i]->isword=1;
                que[tail++]=tmp->next[i];
            }
            else if(tmp==root) tmp->next[i]=root;
            else tmp->next[i]=tmp->fail->next[i];
        }
    }
}
int dp[1005][2005];
int slove(char *str,int len){
    for(int i=0;i<=len;i++) for(int j=0;j<idx;j++) dp[i][j]=inf;
    dp[0][0]=0;
    for(int i=1;i<=len;i++){
        for(int j=0;j<idx;j++){
            if(s[j].isword) continue;
            if(dp[i-1][j]==inf) continue;
            for(int k=0;k<4;k++){
                int r=s[j].next[k]->kind;
                if(s[r].isword) continue;
                dp[i][r]=min(dp[i][r],dp[i-1][j]+(id(str[i-1])!=k));
            }
        }
    }
    int ans=inf;
    for(int i=0;i<idx;i++) ans=min(ans,dp[len][i]);
    return ans==inf?-1:ans;
}
char str[1005];
int main(){
    int n,cas=0;
    while(scanf("%d",&n)!=EOF&&n){
        idx=0;
        Trie *root=NewNode();
        for(int i=0;i<n;i++){
            scanf("%s",str);
            Insert(root,str,strlen(str));
        }
        Bulid_Fail(root);
        scanf("%s",str);
        printf("Case %d: %d\n",++cas,slove(str,strlen(str)));
    }
    return 0;
}

DNA repair问题的更多相关文章

  1. hdu2457:DNA repair

    AC自动机+dp.问改变多少个字符能让目标串不含病毒串.即走过多少步不经过病毒串终点.又是同样的问题. #include<cstdio> #include<cstring> # ...

  2. HDU 2425 DNA repair (AC自动机+DP)

    DNA repair Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total ...

  3. 【POJ3691】 DNA repair (AC自动机+DP)

    DNA repair Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u Description B ...

  4. POJ 3691 &amp; HDU 2457 DNA repair (AC自己主动机,DP)

    http://poj.org/problem?id=3691 http://acm.hdu.edu.cn/showproblem.php?pid=2457 DNA repair Time Limit: ...

  5. POJ 3691 DNA repair (DP+AC自动机)

    DNA repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 4815   Accepted: 2237 Descri ...

  6. 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 ...

  7. poj 3691 DNA repair(AC自己主动机+dp)

    DNA repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 5877   Accepted: 2760 Descri ...

  8. HDU2457 DNA repair —— AC自动机 + DP

    题目链接:https://vjudge.net/problem/HDU-2457 DNA repair Time Limit: 5000/2000 MS (Java/Others)    Memory ...

  9. HDU 2457/POJ 3691 DNA repair AC自动机+DP

    DNA repair Problem Description   Biologists finally invent techniques of repairing DNA that contains ...

随机推荐

  1. 解决svn的working copy locked并且cleanup恢复不能的情况

    产生这种情况大多是因为上次svn命令执行失败且被锁定了. 如果cleanup没有效果的话只好手动删除锁定文件. cd 到svn项目目录下,然后执行如下命令 del lock /q/s 就把锁删掉了.

  2. beanFactoory介绍

  3. 【C#】ContextMenuStrip 右键菜单颜色设置

    有些时候自己想要修改ContexMenuStrip右键菜单的一些背景色之类的,该如何实现呢? 首先: ContextMenuStrip _context = new ContextMenuStrip( ...

  4. Linux Linux程序练习十(网络编程大文件发送)

    //网络编程客户端--大文件传输 #include <stdio.h> #include <stdlib.h> #include <string.h> #inclu ...

  5. [转]Windows网络编程学习-面向连接的编程方式

    直接附上原文链接:windows 网络编程学习-面向连接的编程方式

  6. MySQL基础 - 外键和约束

    在工作中经常会遇到不少不同的观点,比如对于数据库来说那就是是否要设置外键,设置外键的理由自然不必多说,而不设置外键的理由多半为设置外键影响性能,但就目前工作来讲,还没有涉及到因为外键而引发的数据库瓶颈 ...

  7. python package 的两种组织方式

    方式一/package1/ .../__init__.py # 空文件 .../class1.py class Class1: def __init__(self): self.name = &quo ...

  8. DDD 领域驱动设计-谈谈 Repository、IUnitOfWork 和 IDbContext 的实践(转)

    http://www.cnblogs.com/xishuai/p/ddd-repository-iunitofwork-and-idbcontext.html 好久没写 DDD 领域驱动设计相关的文章 ...

  9. 简谈Java的join()方法

    join()是Thread类的一个方法.根据jdk文档的定义: public final void join()throws InterruptedException: Waits for this ...

  10. Caffe学习系列(15):计算图片数据的均值

    图片减去均值后,再进行训练和测试,会提高速度和精度.因此,一般在各种模型中都会有这个操作. 那么这个均值怎么来的呢,实际上就是计算所有训练样本的平均值,计算出来后,保存为一个均值文件,在以后的测试中, ...