POJ 3691 & HDU 2457 DNA repair (AC自動機，DP)

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

題意：

給出N個模式串和一個文本串，問最少修改文本串中多少個字母使得文本串中不包含模式串。

分析：

N個模式串建構AC自動機，然後文本串在AC自動機中走，其中單詞結點不可達。

用dp[i][j]表示文本串第i個字母轉移到AC自動機第j個結點最少修改字母的個數，狀态轉移方程為dp[i][j]=min(dp[i][j],dp[i-1][last]+add)，last表示j的前趨，add為目前點是否修改。由于第i個隻和第i-1個有關，是以可以使用滾動數組來優化空間。

/*
 *
 * Author : fcbruce <[email protected]>
 *
 * 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;
}