小明系列故事――女友的考验 HDU - 4511 AC自动机+简单DP

题意：自己看题目，中文体面。

题解：

把所有不能走的路径放入AC自动机中。

然后DP【i】【j】表示走到 i 这个点，且位于AC自动机 j 这个节点最短距离

然后直接DP即可。注意一点会爆int

 #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, k, x[maxn], y[maxn], buf[];
 
 double cal(int id1, int id2) {
     return sqrt((1.0 * x[id1] - x[id2]) * (1.0 * x[id1] - x[id2]) + (1.0 * y[id1] - y[id2]) * (1.0 * y[id1] - y[id2]));
 }
 
 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(int len) {
         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]++;
     }
 
     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] = ;
             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]);
                 }
         }
     }
 
     double dp[][];
 
     void solve() {
         for (int i = ; i <= n; ++i)
             for (int j = ; j < cnt; ++j)
                 dp[i][j] = INFLL;
         dp[][next[root][]] = ;
         for (int i = ; i <= n; ++i) {
             for (int j = ; j < cnt; ++j) {
                 if (dp[i][j] == INFLL) continue;
                 for (int k = i + ; k <= n; ++k) {
                     int idx = next[j][k];
                     if (End[idx]) continue;
                     dp[k][idx] = min(dp[k][idx], dp[i][j] + cal(i, k));
                 }
             }
         }
         double ans = INFLL;
         for (int i = ; i < cnt; ++i) ans = min(ans, dp[n][i]);
         if (ans == INFLL) printf("Can not be reached!\n");
         else printf("%.2f\n", 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;
         for (int i = ; i <= n; ++i) sffi(x[i], y[i]);
         ac.init();
         for (int i = ; i < m; ++i) {
             sfi(k);
             for (int j = ; j < k; ++j) sfi(buf[j]);
             ac.insert(k);
         }
         ac.build();
         ac.solve();
     }
     return ;
 }