题目链接：hdu 2457 DNA repair

题目大意：给定一些DNA序列，表示带有疾病，现在给定一个DNA序列，要求修改最少的位置，使得DNA不带有疾病的

片段。

解题思路：AC自动机+DP，先将DNA片段建立AC自动机，然后在AC自动机上进行dp，dp[i][j]表示长度为i移动到j节点

修改了最少的步数。每次走到边如果和字符串不同，权值即为1；相同则为0。单词节点不能向后转移。

#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;

const int maxn = 1005;
const int sigma_size = 4;
const int inf = 0x3f3f3f3f;
const char sign[5] = "ACGT";

struct Aho_Corasick {
    int sz, g[maxn][sigma_size];
    int tag[maxn], fail[maxn], last[maxn];

    int dp[maxn][maxn];

    void init();
    int idx(char ch);
    void insert(char* str, int k);
    void getFail();
    void match(char* str);
    void put(int x, int y);
    int solve(char* s);
}AC;

int N;
char s[maxn];

int main () {
    int cas = 1;
    while (scanf("%d", &N) == 1 && N) {
        AC.init();
        for (int i = 0; i < N; i++) {
            scanf("%s", s);
            AC.insert(s, i+1);
        }
        scanf("%s", s);
        printf("Case %d: %d\n", cas++, AC.solve(s));
    }
    return 0;
}

int Aho_Corasick::solve(char* s) {
    getFail();

    int n = strlen(s);
    memset(dp, inf, sizeof(dp));
    dp[0][0] = 0;

    for (int k = 0; k < n; k++) {
        for (int i = 0; i < sz; i++) {
            if (dp[k][i] == inf || tag[i] || last[i])
                continue;

            for (int j = 0; j < 4; j++) {
                int u = i;
                while (u && g[u][j] == 0)
                    u = fail[u];
                u = g[u][j];

                int d = (sign[j] == s[k] ? 0 : 1);
                if (dp[k+1][u] > dp[k][i] + d)
                    dp[k+1][u] = dp[k][i] + d;
            }
        }
    }

    int ans = inf;
    for (int i = 0; i < sz; i++) {
        if (tag[i] || last[i])
            continue;
        ans = min(ans, dp[n][i]);
    }
    return ans == inf ? -1 : ans;
}

void Aho_Corasick::init() {
    sz = 1;
    tag[0] = 0;
    memset(g[0], 0, sizeof(g[0]));
}

int Aho_Corasick::idx(char ch) {
    if (ch == ‘A‘)
        return 0;
    if (ch == ‘C‘)
        return 1;
    if (ch == ‘G‘)
        return 2;
    return 3;
}

void Aho_Corasick::put(int x, int y) {
}

void Aho_Corasick::insert(char* str, int k) {
    int u = 0, n = strlen(str);

    for (int i = 0; i < n; i++) {
        int v = idx(str[i]);
        if (g[u][v] == 0) {
            tag[sz] = 0;
            memset(g[sz], 0, sizeof(g[sz]));
            g[u][v] = sz++;
        }
        u = g[u][v];
    }
    tag[u] = 1;
}

void Aho_Corasick::match(char* str) {
    int n = strlen(str), u = 0;
    for (int i = 0; i < n; i++) {
        int v = idx(str[i]);
        while (u && g[u][v] == 0)
            u = fail[u];

        u = g[u][v];

        if (tag[u])
            put(i, u);
        else if (last[u])
            put(i, last[u]);
    }
}

void Aho_Corasick::getFail() {
    queue<int> que;

    for (int i  = 0; i < sigma_size; i++) {
        int u = g[0][i];
        if (u) {
            fail[u] = last[u] = 0;
            que.push(u);
        }
    }

    while (!que.empty()) {
        int r = que.front();
        que.pop();

        for (int i = 0; i < sigma_size; i++) {
            int u = g[r][i];

            if (u == 0) {
                g[r][i] = g[fail[r]][i];
                continue;
            }

            que.push(u);
            int v = fail[r];
            while (v && g[v][i] == 0)
                v = fail[v];

            fail[u] = g[v][i];
            last[u] = tag[fail[u]] ? fail[u] : last[fail[u]];
        }
    }
}

hdu 2457 DNA repair(AC自动机+DP)