Resource Archiver HDU - 3247 AC自动机+BFS+状压

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("******\n")
 #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 = ;
 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 const int maxn = 1e6 + ;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = 1e9 + ;
 int n, m;
 char buf[maxn];

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

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

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

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

     void build() {
         queue<int> Q;
         fail[root] = root;
         for (int i = ; i < ; i++)
             if (next[root][i] == -) 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]] == -) End[now] = -;
             else End[now] |= End[fail[now]];
             for (int i = ; i < ; i++)
                 if (next[now][i] == -) next[now][i] = next[fail[now]][i];
                 else {
                     fail[next[now][i]] = next[fail[now]][i];
                     Q.push(next[now][i]);
                 }
         }
     }

     int mp[][], dis[], pos[], num, dp[][];

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

     int solve() {
         pos[] = , num = ;
         for (int i = ; i < cnt; ++i) if (End[i] > ) pos[num++] = i;
         for (int i = ; 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("\n");
 //        }

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

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

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