http://poj.org/problem?id=3691
http://acm.hdu.edu.cn/showproblem.php?pid=2457
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DNA repair
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 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 |
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 <[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;
}