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POJ 3691 & HDU 2457 DNA repair (AC自动机，DP)

2024-08-02 21:50:37   217人阅读

http://poj.org/problem?id=3691

http://acm.hdu.edu.cn/showproblem.php?pid=2457

DNA repair
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 5690 Accepted: 2669

Description

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

Source

2008 Asia Hefei Regional Contest Online by USTC


题意：

给出N个模式串和一个文本串，问最少修改文本串中多少个字母使得文本串中不包含模式串。

分析：

N个模式串构建AC自动机，然后文本串在AC自动机中走，其中单词结点不可达。

用dp[i][j]表示文本串第i个字母转移到AC自动机第j个结点最少修改字母的个数，状态转移方程为dp[i][j]=min(dp[i][j],dp[i-1][last]+add)，last表示j的前趋，add为当前点是否修改。由于第i个只和第i-1个有关，所以可以使用滚动数组来优化空间。


/*
 *
 * Author : fcbruce <fcbruce8964@gmail.com>
 *
 * Time : Tue 18 Nov 2014 11:17:49 AM CST
 *
 */
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10

#ifdef _WIN32
  #define lld "%I64d"
#else
  #define lld "%lld"
#endif

#define maxm 
#define maxn 1024

using namespace std;

int q[maxn];

const int maxsize = 4;
struct Acauto
{
  int ch[maxn][maxsize];
  bool val[maxn];
  int last[maxn],nex[maxn];
  int sz;
  int dp[2][maxn];

  Acauto()
  {
    memset(ch[0],0,sizeof ch[0]);
    val[0]=false;
    sz=1;
  }

  void clear()
  {
    memset(ch[0],0,sizeof ch[0]);
    val[0]=false;
    sz=1;
  }

  int idx(const char c)
  {
    if (c=='A') return 0;
    if (c=='T') return 1;
    if (c=='C') return 2;
    return 3;
  }

  void insert(const char *s)
  {
    int u=0;
    for (int i=0;s[i]!='\0';i++)
    {
      int c=idx(s[i]);
      if (ch[u][c]==0)
      {
        memset(ch[sz],0,sizeof ch[sz]);
        val[sz]=false;
        ch[u][c]=sz++;
      }
      u=ch[u][c];
    }
    val[u]=true;
  }

  void get_fail()
  {
    int f=0,r=-1;
    nex[0]=0;
    for (int c=0;c<maxsize;c++)
    {
      int u=ch[0][c];
      if (u!=0)
      {
        nex[u]=0;
        q[++r]=u;
        last[u]=0;
      }
    }

    while (f<=r)
    {
      int x=q[f++];
      for (int c=0;c<maxsize;c++)
      {
        int u=ch[x][c];
        if (u==0)
        {
          ch[x][c]=ch[nex[x]][c];
          continue;
        }
        q[++r]=u;
        int v=nex[x];
        nex[u]=ch[v][c];
        val[u]|=val[nex[u]];
      }
    }
  }

  int DP(const char *T)
  {
    memset(dp,0x3f,sizeof dp);
    dp[0][0]=0;
    int x=1;
    for (int i=0;T[i]!='\0';i++,x^=1)
    {
      memset(dp[x],0x3f,sizeof dp[x]);
      int c=idx(T[i]);
      for (int j=0;j<sz;j++)
      {
        if (dp[x^1][j]==INF) continue;
        for (int k=0;k<4;k++)
        {
          if (val[ch[j][k]]) continue;
          int add=k==c?0:1;
          dp[x][ch[j][k]]=min(dp[x][ch[j][k]],dp[x^1][j]+add);
        }
      }
    }

    int MIN=INF;
    for (int i=0;i<sz;i++)
      MIN=min(MIN,dp[x^1][i]);
    if (MIN==INF) MIN=-1;
    return MIN;
  }
}acauto;

char DNA[1024];

int main()
{
#ifdef FCBRUCE
  freopen("/home/fcbruce/code/t","r",stdin);
#endif // FCBRUCE

  int n,__=0;
  
  while (scanf("%d",&n),n!=0)
  {
    acauto.clear();
    for (int i=0;i<n;i++)
    {
      scanf("%s",DNA);
      acauto.insert(DNA);
    }

    acauto.get_fail();

    scanf("%s",DNA);

    printf("Case %d: %d\n",++__,acauto.DP(DNA));
  }


  return 0;
}


POJ 3691 & HDU 2457 DNA repair (AC自动机，DP)


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