hdu 2457 DNA repair

          AC自动机+DP。按着自动机跑，（其实是生成新的满足题目要求的串，然后找改变最少的。）但是不能跑到是单词的地方，如果跑到单词的话那么说明改变后的串含有病毒了，不满足题意。然后就是应该怎么跑的问题了，现在我们从自动机的根节点开始跑，如果跑到下一个节点和当前串的字母不一样的话，那么当前位置生成的串是和原串在该位置是有差异的，dp+1，否者的话dp不变。所以dp[ i ][ j ]表示的是匹配到当前匹配串的位置时，跑到自动机的 j 节点需要改变的最少字母数。

 

#include <algorithm>

#include <iostream>

#include <sstream>

#include <cstdlib>

#include <climits>

#include <cstring>

#include <cstdio>

#include <string>

#include <vector>

#include <cctype>

#include <queue>

#include <cmath>

#include <set>

#include <map>

#define CLR(a, b) memset(a, b, sizeof(a))

using namespace std;



const int MAX_NODE = 22 * 55 * 2;

const int INF = 0x3f3f3f3f;

const int CHILD_NUM = 4;

const int N = 1010;



class ACAutomaton

{

private:

    int chd[MAX_NODE][CHILD_NUM];

    int dp[N][MAX_NODE];

    int fail[MAX_NODE];

    bool val[MAX_NODE];

    int Q[MAX_NODE];

    int ID[128];

    int sz;

public:

    void Initialize()

    {

        fail[0] = 0;

        ID['A'] = 0;ID['G'] = 1;

        ID['C'] = 2;ID['T'] = 3;

    }

    void Reset()

    {

        CLR(chd[0] , 0);sz = 1;

    }

    void Insert(char *a)

    {

        int p = 0;

        for ( ; *a ; a ++)

        {

            int c = ID[*a];

            if (!chd[p][c])

            {

                CLR(chd[sz] , 0);

                val[sz] = false;

                chd[p][c] = sz ++;

            }

            p = chd[p][c];

        }

        val[p] = true;

    }

    void Construct()

    {

        int *s = Q , *e = Q;

        for (int i = 0 ; i < CHILD_NUM ; i ++)

        {

            if (chd[0][i])

            {

                fail[ chd[0][i] ] = 0;

                *e ++ = chd[0][i];

            }

        }

        while (s != e)

        {

            int u = *s++;

            for (int i = 0 ; i < CHILD_NUM ; i ++)

            {

                int &v = chd[u][i];

                if (v)

                {

                    *e ++ = v;

                    fail[v] = chd[ fail[u] ][i];

                    val[v] |= val[fail[v]];

                }

                else

                {

                    v = chd[ fail[u] ][i];

                }

            }

        }

    }

    int Work(char *ch)

    {

        int len, S, T, ret;

        len = strlen(ch);

        CLR(dp, INF);dp[0][0] = 0;

        for(int i = 0; i < len; i ++)

            for(int j = 0; j < sz; j ++)

            {

                if(val[j]) continue;

                if(dp[i][j] == INF) continue;

                for(int k = 0; k < 4; k ++)

                {

                    T = chd[j][k];

                    if(val[T]) continue;

                    dp[i + 1][T] = min(dp[i + 1][T], dp[i][j] + (ID[ch[i]] != k));

                }

        }ret = INF;

        for(int i = 0; i < sz; i ++)

        {

            ret = min(ret, dp[len][i]);

        }

        return ret == INF ? -1 : ret;

    }

} AC;



char ch[N];



int main()

{

    //freopen("input.txt", "r", stdin);

    AC.Initialize();

    int n, t, cas = 1;

    while (scanf("%d", &n), n)

    {

        AC.Reset();

        for (int i = 0 ; i < n ; i ++)

        {

            char temp[55];

            scanf("%s", temp);

            AC.Insert(temp);

        }

        scanf("%s", ch);

        AC.Construct();

        printf("Case %d: %d\n", cas ++, AC.Work(ch));

    }

    return 0;

}