AC自动机+DP HDOJ 2457 DNA repair（DNA修复）

题目链接

题意：

　　给n串有疾病的DNA序列，现有一串DNA序列，问最少修改几个DNA，能使新的DNA序列不含有疾病的DNA序列。

思路：

　　构建AC自动机，设定end结点，dp[i][j]表示长度i的前缀串走到自动机的j点最少需要修改几个DNA。状态转移方程。那么只要转移到下一个的DNA不是end结点就能转移，如果下一个DNA不和原序列不一样就+1。

#include <bits/stdc++.h>const int N = 50 + 5;
const int M = 1000 + 5;
const int INF = 0x3f3f3f3f;
struct AC {
    static const int NODE = N * 20;
    static const int SIZE = 4;
    int ch[NODE][SIZE], fail[NODE], end[NODE];
    int sz;    void init();
    int idx(char c);
    void insert(char *s);
    void get_fail();
}ac;char s[25];
char str[M];
int dp[M][AC::NODE];int solve() {
    int len = strlen (str);
    memset (dp, INF, sizeof (dp));
    dp[0][0] = 0;    for (int i=1; i<=len; ++i) {
        for (int j=0; j<ac.sz; ++j) {
            if (dp[i-1][j] == INF) continue;
            for (int k=0; k<4; ++k) {
                if (!ac.end[ac.ch[j][k]]) {
                    dp[i][ac.ch[j][k]] = std::min (dp[i][ac.ch[j][k]], dp[i-1][j] + (ac.idx (str[i-1]) != k));
                }
            }
        }
    }    int ret = INF;
    for (int i=0; i<ac.sz; ++i) {
        if (!ac.end[i]) {
            ret = std::min (ret, dp[len][i]);
        }
    }
    if (ret == INF) ret = -1;
    return ret;
}int main() {
    int n, cas = 0;
    while (scanf ("%d", &n) == 1 && n) {
        ac.init ();
        for (int i=0; i<n; ++i) {
            scanf ("%s", s);
            ac.insert (s);
        }
        ac.get_fail ();
        scanf ("%s", str);        printf ("Case %d: %d\n", ++cas, solve ());
    }
    return 0;
}void AC::init() {
    memset (ch[0], 0, sizeof (ch[0]));
    sz = 1;
}int AC::idx(char c) {
    if (c == 'A') {
        return 0;
    } else if (c == 'G') {
        return 1;
    } else if (c == 'C') {
        return 2;
    } else {
        return 3;
    }
}void AC::insert(char *s) {
    int u = 0;
    for (int c, i=0; s[i]; ++i) {
        c = idx (s[i]);
        if (!ch[u][c]) {
            memset (ch[sz], 0, sizeof (ch[sz]));
            end[sz] = 0;
            ch[u][c] = sz++;
        }
        u = ch[u][c];
    }
    end[u] = 1;
}void AC::get_fail() {
    fail[0] = 0;
    std::queue<int> que;
    for (int c=0; c<SIZE; ++c) {
        int u = ch[0][c];
        if (u) {
            fail[u] = 0;
            //last[u] = 0;
            que.push (u);
        }
    }
    while (!que.empty ()) {
        int r = que.front ();
        que.pop ();
        for (int c=0; c<SIZE; ++c) {
            int &u = ch[r][c];
            if (!u) {
                u = ch[fail[r]][c];
            } else {
                int v = fail[r];
                while (v && !ch[v][c]) v = fail[v];
                fail[u] = ch[v][c];
                end[u] |= end[fail[u]];
                //last[u] = end[fail[u]] ? fail[u] : last[fail[u]];
                que.push (u);
            }
        }
    }
}