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。
1 #include <set> 2 #include <map> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <cstdio> 8 #include <string> 9 #include <vector> 10 #include <cstring> 11 #include <iostream> 12 #include <algorithm> 13 #include <unordered_map> 14 15 #define pi acos(-1.0) 16 #define eps 1e-9 17 #define fi first 18 #define se second 19 #define rtl rt<<1 20 #define rtr rt<<1|1 21 #define bug printf("******\n") 22 #define mem(a, b) memset(a,b,sizeof(a)) 23 #define name2str(x) #x 24 #define fuck(x) cout<<#x" = "<<x<<endl 25 #define sfi(a) scanf("%d", &a) 26 #define sffi(a, b) scanf("%d %d", &a, &b) 27 #define sfffi(a, b, c) scanf("%d %d %d", &a, &b, &c) 28 #define sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d) 29 #define sfL(a) scanf("%lld", &a) 30 #define sffL(a, b) scanf("%lld %lld", &a, &b) 31 #define sfffL(a, b, c) scanf("%lld %lld %lld", &a, &b, &c) 32 #define sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d) 33 #define sfs(a) scanf("%s", a) 34 #define sffs(a, b) scanf("%s %s", a, b) 35 #define sfffs(a, b, c) scanf("%s %s %s", a, b, c) 36 #define sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d) 37 #define FIN freopen("../in.txt","r",stdin) 38 #define gcd(a, b) __gcd(a,b) 39 #define lowbit(x) x&-x 40 #define IO iOS::sync_with_stdio(false) 41 42 43 using namespace std; 44 typedef long long LL; 45 typedef unsigned long long ULL; 46 const ULL seed = 13331; 47 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL; 48 const int maxn = 1e6 + 7; 49 const int maxm = 8e6 + 10; 50 const int INF = 0x3f3f3f3f; 51 const int mod = 1e9 + 7; 52 int n, m; 53 char buf[maxn]; 54 55 struct Aho_Corasick { 56 int next[60010][2], fail[60010], End[60010]; 57 int root, cnt; 58 59 int newnode() { 60 for (int i = 0; i < 2; i++) next[cnt][i] = -1; 61 End[cnt++] = 0; 62 return cnt - 1; 63 } 64 65 void init() { 66 cnt = 0; 67 root = newnode(); 68 } 69 70 void insert(char buf[], int id) { 71 int len = strlen(buf); 72 int now = root; 73 for (int i = 0; i < len; i++) { 74 if (next[now][buf[i] - '0'] == -1) next[now][buf[i] - '0'] = newnode(); 75 now = next[now][buf[i] - '0']; 76 } 77 End[now] = id; 78 } 79 80 void build() { 81 queue<int> Q; 82 fail[root] = root; 83 for (int i = 0; i < 2; i++) 84 if (next[root][i] == -1) next[root][i] = root; 85 else { 86 fail[next[root][i]] = root; 87 Q.push(next[root][i]); 88 } 89 while (!Q.empty()) { 90 int now = Q.front(); 91 Q.pop(); 92 if (End[fail[now]] == -1) End[now] = -1; 93 else End[now] |= End[fail[now]]; 94 for (int i = 0; i < 2; i++) 95 if (next[now][i] == -1) next[now][i] = next[fail[now]][i]; 96 else { 97 fail[next[now][i]] = next[fail[now]][i]; 98 Q.push(next[now][i]); 99 } 100 } 101 } 102 103 int mp[15][15], dis[60010], pos[15], num, dp[15][1055]; 104 105 void bfs(int x) { 106 mem(dis, -1); 107 dis[pos[x]] = 0; 108 queue<int> q; 109 q.push(pos[x]); 110 while (!q.empty()) { 111 int now = q.front(); 112 q.pop(); 113 for (int i = 0; i < 2; ++i) { 114 int idx = next[now][i]; 115 if (End[idx] == -1 || dis[idx] >= 0) continue; 116 dis[idx] = dis[now] + 1, q.push(idx); 117 } 118 } 119 for (int i = 0; i < num; ++i) mp[x][i] = dis[pos[i]]; 120 } 121 122 int solve() { 123 pos[0] = 0, num = 1; 124 for (int i = 0; i < cnt; ++i) if (End[i] > 0) pos[num++] = i; 125 for (int i = 0; i < num; ++i) bfs(i); 126 127 // for (int i = 0; i < num; ++i) { 128 // for (int j = 0; j < num; ++j) { 129 // printf("%d ", mp[i][j]); 130 // } 131 // printf("\n"); 132 // } 133 134 135 for (int i = 0; i < num; ++i) 136 for (int j = 0; j < (1 << n); ++j) 137 dp[i][j] = INF; 138 dp[0][0] = 0; 139 for (int status = 0; status < (1 << n); ++status) { 140 for (int i = 0; i < num; ++i) { 141 if (dp[i][status] == INF) continue; 142 for (int j = 0; j < num; ++j) { 143 if (mp[i][j] == -1 || (status | End[pos[j]]) == status) continue; 144 dp[j][status | End[pos[j]]] = min(dp[j][status | End[pos[j]]], dp[i][status] + mp[i][j]); 145 } 146 } 147 } 148 int ans = INF; 149 for (int i = 0; i < num; ++i) ans = min(ans, dp[i][(1 << n) - 1]); 150 return ans; 151 } 152 153 void debug() { 154 for (int i = 0; i < cnt; i++) { 155 printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]); 156 for (int j = 0; j < 26; j++) printf("%2d", next[i][j]); 157 printf("]\n"); 158 } 159 } 160 } ac; 161 162 int main() { 163 // FIN; 164 while (~sffi(n, m)) { 165 if (n == 0 && m == 0) break; 166 ac.init(); 167 for (int i = 0; i < n; ++i) { 168 sfs(buf); 169 ac.insert(buf, (1 << i)); 170 } 171 for (int i = 0; i < m; ++i) { 172 sfs(buf); 173 ac.insert(buf, -1); 174 } 175 ac.build(); 176 printf("%d\n", ac.solve()); 177 } 178 return 0; 179 }
