题目链接：poj 1699 Best Sequence

题目大意；给定N个DNA序列，问说最少多长的字符串包含所有序列。

解题思路：AC自动机+状压DP，先对字符串构造AC自动机，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]];
        }
    }
}

poj 1699 Best Sequence(AC自动机+状压DP)