HDOJ 2457 DNA repair (AC自动机+DP)

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题意

给出一些子串和一个DNA串，求最少需要变换多少个字符可以让这个DNA不包含任意一个子串。

思路

dp[i][j]表示在字符串i处在trie图的节点j处时的最小改变数。
然后对字符串和每个节点都枚举递推一遍就行了。

代码

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define LL long long#define Lowbit(x) ((x)&(-x))#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1|1#define MP(a, b) make_pair(a, b)const int INF = 0x3f3f3f3f;const int MOD = 1000000007;const int maxn = 200010;const double eps = 1e-8;const double PI = acos(-1.0);typedef pair<int, int> pii;struct Trie{    int next[1010][4], fail[1010];    bool end[1010];    int root, L;    int newnode()    {        for (int i = 0; i < 4; i++)            next[L][i] = -1;        end[L++] = false;        return L - 1;    }    void init()    {        L = 0;        root = newnode();    }    void insert(char s[])    {        int len = strlen(s);        int now = root;        for (int i = 0; i < len; i++)        {            if (next[now][getch(s[i])] == -1)                next[now][getch(s[i])] = newnode();            now = next[now][getch(s[i])];        }        end[now] = true;    }    void build()    {        queue<int> Q;        fail[root] = root;        for (int i = 0; i < 4; i++)            if (next[root][i] == -1)                next[root][i] = root;            else            {                fail[next[root][i]] = root;                Q.push(next[root][i]);            }        while (!Q.empty())        {            int now = Q.front();            Q.pop();            if (end[fail[now]] == true)                end[now] = true;            for (int i = 0; i < 4; i++)                if (next[now][i] == -1)                    next[now][i] = next[fail[now]][i];                else                {                    fail[next[now][i]] = next[fail[now]][i];                    Q.push(next[now][i]);                }        }    }//    Matrix getMatrix()//    {//        Matrix res = Matrix(L+1);//        for (int i = 0; i < L; i++)//            for (int j = 0; j < 26; j++)//                if (end[next[i][j]] == false)//                    res.mat[i][next[i][j]]++;//        for (int i = 0; i < L + 1; i++)//            res.mat[i][L] = 1;//        return res;//    }    int getch(char ch)    {        if (ch == 'A') return 0;        if (ch == 'C') return 1;        if (ch == 'G') return 2;        if (ch == 'T') return 3;    }    int dp[1010][1010];    int DP(char buf[])    {        int len = strlen(buf);        memset(dp, INF, sizeof(dp));        dp[0][root] = 0;        for (int i = 0; i < len; i++)            for (int j = 0; j < L; j++)                if (dp[i][j] != INF)                {                    for (int k = 0; k < 4; k++)                    {                        int now = next[j][k];                        if (end[now]) continue;                        int temp;                        if (getch(buf[i]) == k) temp = dp[i][j];                        else temp = dp[i][j] + 1;                        dp[i+1][now] = min(dp[i+1][now], temp);                    }                }        int ans = INF;        for (int i = 0; i < L; i++)            ans = min(ans, dp[len][i]);        if (ans == INF) return -1;        return ans;    }};char buf[1010];Trie ac;int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int n, ncase = 1;    while (scanf("%d", &n) && n)    {        ac.init();        for (int i = 0; i < n; i++)        {            scanf("%s", buf);            ac.insert(buf);        }        ac.build();        scanf("%s", buf);        printf("Case %d: %d\n", ncase++, ac.DP(buf));    }    return 0;}

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