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. Android数据存储(一)----SharedPreferences详解

    一.Android数据的存储方式: Android系统一共提供了四种数据存储方式.分别是:SharePreference.SQLite.Content Provider和File:此外还有一种网络存储 ...

  2. smarty缓存控制

    第一步初始化配置文件中设置 如果当前访问的模板有缓存就不需要连接数据库那些代码了,如果要模板局部不缓存,要写在iscache外,模板中用{nocache}

  3. 没有什么好神秘的: wait_on_page_bit

    文件系统中经常会有wait_on_page_bit函数的封装,比如f2fs中就会有如下的代码: 1431 void f2fs_wait_on_page_writeback(struct page *p ...

  4. TabControl的SelectionChanged事件

    DataGrid作为TabControl控件的TabItem的content元素. 当操作DataGrid的不同cell时,会引发了TabControl的SelectionChanged事件的问题. ...

  5. 第二章 时间控件(DateTime Picker)

    这家伙太懒了,碰到问题才写博文,嘿嘿. 好了进入正题,二话不说,先放地址: 中文:http://www.bootcss.com/p/bootstrap-datetimepicker/index.htm ...

  6. 【WPF】WPF通过RelativeSource绑定父控件的属性

    1.后台代码实现绑定父控件的属性 RelativeSource rs = new RelativeSource(RelativeSourceMode.FindAncestor); //设定为离自己控件 ...

  7. [tomcat7源码学习]初始化之catalina.home和catalina.base(转)

    我们在代码中为了获取某个配置文件路径下的文件经常会这么写 String tomcatPath = System.getProperty("catalina.home") + &qu ...

  8. Jquery 将表单序列化为Json对象

    大家知道Jquery中有serialize方法,可以将表单序列化为一个“&”连接的字符串,但却没有提供序列化为Json的方法.不过,我们可以写一个插件实现. 我在网上看到有人用替换的方法,先用 ...

  9. 用python简单处理图片(5):图像直方图

    我们先来看两个函数reshape和flatten: 假设我们先生成一个一维数组: vec=np.arange(15) print vec 显示为: [ 0 1 2 3 4 5 6 7 8 9 10 1 ...

  10. Ubuntu Navicat正版永久使用方法

    最近技安不再带自己的mac book来公司工作了,公司多出来一个台式机,配置还挺高.于是乎就拿过来用了. 装上了Ubuntu14.04 LTS版,正常的开发工具如vagrant,vitualbox,s ...