Resource Archiver HDU

题意：

给出n个资源串，m个病毒串，现在要如何连接资源串使得不含病毒串（可以重叠，样例就是重叠的）。

题解：

这题的套路和之前的很不同了，之前的AC自动机+DP的题目一般都是通过teir图去转移，

这题有点小不同，认真分析这一题，我们可以知道n个资源串，m个病毒串在AC自动机的上面是哪一个节点

所以可以根据teir图去处理出每一个资源串的节点不经过病毒串节点的最短距离，这个可以通过BFS实现。

这个一开始要处理AC自动机上的root节点，毕竟是从root节点开始走的。

处理出了每个点到其他点的最短距离之后，就变成了一个TSP问题，必须经过几个点的最短距离，状压写即可。

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>

#define  pi    acos(-1.0)
#define  eps   1e-9
#define  fi    first
#define  se    second
#define  rtl   rt<<1
#define  rtr   rt<<1|1
#define  bug                printf("******
")
#define  mem(a, b)          memset(a,b,sizeof(a))
#define  name2str(x)        #x
#define  fuck(x)            cout<<#x" = "<<x<<endl
#define  sfi(a)             scanf("%d", &a)
#define  sffi(a, b)         scanf("%d %d", &a, &b)
#define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
#define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define  sfL(a)             scanf("%lld", &a)
#define  sffL(a, b)         scanf("%lld %lld", &a, &b)
#define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
#define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
#define  sfs(a)             scanf("%s", a)
#define  sffs(a, b)         scanf("%s %s", a, b)
#define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
#define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
#define  FIN                freopen("../in.txt","r",stdin)
#define  gcd(a, b)          __gcd(a,b)
#define  lowbit(x)          x&-x
#define  IO                 iOS::sync_with_stdio(false)


using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const ULL seed = 13331;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int maxn = 1e6 + 7;
const int maxm = 8e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n, m;
char buf[maxn];

struct Aho_Corasick {
    int next[60010][2], fail[60010], End[60010];
    int root, cnt;

    int newnode() {
        for (int i = 0; i < 2; i++) next[cnt][i] = -1;
        End[cnt++] = 0;
        return cnt - 1;
    }

    void init() {
        cnt = 0;
        root = newnode();
    }

    void insert(char buf[], int id) {
        int len = strlen(buf);
        int now = root;
        for (int i = 0; i < len; i++) {
            if (next[now][buf[i] - '0'] == -1) next[now][buf[i] - '0'] = newnode();
            now = next[now][buf[i] - '0'];
        }
        End[now] = id;
    }

    void build() {
        queue<int> Q;
        fail[root] = root;
        for (int i = 0; i < 2; 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]] == -1) End[now] = -1;
            else End[now] |= End[fail[now]];
            for (int i = 0; i < 2; 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]);
                }
        }
    }

    int mp[15][15], dis[60010], pos[15], num, dp[15][1055];

    void bfs(int x) {
        mem(dis, -1);
        dis[pos[x]] = 0;
        queue<int> q;
        q.push(pos[x]);
        while (!q.empty()) {
            int now = q.front();
            q.pop();
            for (int i = 0; i < 2; ++i) {
                int idx = next[now][i];
                if (End[idx] == -1 || dis[idx] >= 0) continue;
                dis[idx] = dis[now] + 1, q.push(idx);
            }
        }
        for (int i = 0; i < num; ++i) mp[x][i] = dis[pos[i]];
    }

    int solve() {
        pos[0] = 0, num = 1;
        for (int i = 0; i < cnt; ++i) if (End[i] > 0) pos[num++] = i;
        for (int i = 0; i < num; ++i) bfs(i);

//        for (int i = 0; i < num; ++i) {
//            for (int j = 0; j < num; ++j) {
//                printf("%d ", mp[i][j]);
//            }
//            printf("
");
//        }


        for (int i = 0; i < num; ++i)
            for (int j = 0; j < (1 << n); ++j)
                dp[i][j] = INF;
        dp[0][0] = 0;
        for (int status = 0; status < (1 << n); ++status) {
            for (int i = 0; i < num; ++i) {
                if (dp[i][status] == INF) continue;
                for (int j = 0; j < num; ++j) {
                    if (mp[i][j] == -1 || (status | End[pos[j]]) == status) continue;
                    dp[j][status | End[pos[j]]] = min(dp[j][status | End[pos[j]]], dp[i][status] + mp[i][j]);
                }
            }
        }
        int ans = INF;
        for (int i = 0; i < num; ++i) ans = min(ans, dp[i][(1 << n) - 1]);
        return ans;
    }

    void debug() {
        for (int i = 0; i < cnt; i++) {
            printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
            for (int j = 0; j < 26; j++) printf("%2d", next[i][j]);
            printf("]
");
        }
    }
} ac;

int main() {
   // FIN;
    while (~sffi(n, m)) {
        if (n == 0 && m == 0) break;
        ac.init();
        for (int i = 0; i < n; ++i) {
            sfs(buf);
            ac.insert(buf, (1 << i));
        }
        for (int i = 0; i < m; ++i) {
            sfs(buf);
            ac.insert(buf, -1);
        }
        ac.build();
        printf("%d
", ac.solve());
    }
    return 0;
}