poj 1699 Best Sequence(AC自己主动机+如压力DP)

题目大意；给定N个DNA序列，问说最少多长的字符串包括全部序列。
解题思路：AC自己主动机+状压DP，先对字符串构造AC自己主动机。然后在dp[s][i]表示匹配了s。移动到节点i时候的最短步数。
#include <cstdio>

#include <cstring>

#include <queue>

#include <vector>

#include <iostream>

#include <algorithm>

using namespace std;

typedef pair<int,int> pii;

const int maxn = 205;

const int sigma_size = 4;

const int inf = 0x3f3f3f3f;

struct Aho_Corasick {

    int sz, g[maxn][sigma_size];

    int tag[maxn], fail[maxn], last[maxn];

    int dp[maxn][1030];

    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();

}AC;

int N;

char w[30];

int main () {

    int cas;

    scanf("%d", &cas);

    while (cas--) {

        AC.init();

        scanf("%d", &N);

        for (int i = 0; i < N; i++) {

            scanf("%s", w);

            AC.insert(w, i);

        }

        printf("%d\n", AC.solve());

    }

    return 0;

}

int Aho_Corasick::solve() {

    getFail();

    memset(dp, inf, sizeof(dp));

    dp[0][0] = 0;

    queue<pii> que;

    que.push(make_pair(0, 0));

    while(!que.empty()) {

        int u = que.front().first;

        int s = que.front().second;

        que.pop();

        for (int i = 0; i < 4; i++) {

            int k = u;

            while (k && g[k][i] == 0)

                k = fail[k];

            k = g[k][i];

            int ss = s | tag[k];

            if (dp[k][ss] > dp[u][s] + 1) {

                dp[k][ss] = dp[u][s] + 1;

                que.push(make_pair(k, ss));

                if (ss == (1<<N)-1)

                    return dp[k][ss];

            }

        }

    }

    return 0;

}

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<<k);

}

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];

            tag[u] |= tag[fail[u]];

            //last[u] = tag[fail[u]] ?

fail[u] : last[fail[u]];

        }

    }

}