[hdu5164]ac自动机

中文题目：http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=563&pid=1003

首先贴一下bc的题解：

首先我们考虑m=1的情况。给定两个数组A={a1,a2,…,an}和B={b1,b2,…,bk}，问B在A中出现了几次。令ci=ai+1ai,1≤i<n，同样令di=bi+1bi,1≤i<k，那么上述问题可以转化为ci和di的模式匹配问题，这个正确性显然，也没有什么好证明的。于是对于m=1的情况只有用个kmp即可搞定。

现在考虑m>1的情况，我们考虑用ac自动机来做。考虑到字符集并不是普通的数字，而是一个分数，我们不放搞个分数类，然后用map存转移边。用m个模式串（Bob的序列）建好自动机之后，把文本串（Alice的序列）在自动机上跑一遍即可统计出答案。

事实上，这个题的精华就在于序列变换，将问题转化为多模板匹配问题。而且由于转化后的序列有二维状态，可以用map映射重新为状态分配一个值（类似离散）。另外由于字符集比较大，所以ac自动机里面需要一点小小的修改，原来的二维数组的第二维需用map，另外需要同时记录所有儿子的字符值。原来的模板只需要小改就可以适应这种情况了。详见代码：

 #pragma comment(linker, "/STACK:10240000,10240000")

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <map>
 #include <queue>
 #include <deque>
 #include <cmath>
 #include <vector>
 #include <ctime>
 #include <cctype>
 #include <set>
 #include <bitset>
 #include <functional>
 #include <numeric>
 #include <stdexcept>
 #include <utility>

 using namespace std;

 #define mem0(a) memset(a, 0, sizeof(a))
 #define mem_1(a) memset(a, -1, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define define_m int m = (l + r) >> 1
 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 #define all(a) (a).begin(), (a).end()
 #define lowbit(x) ((x) & (-(x)))
 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 #define pchr(a) putchar(a)
 #define pstr(a) printf("%s", a)
 #define sstr(a) scanf("%s", a)
 #define sint(a) scanf("%d", &a)
 #define sint2(a, b) scanf("%d%d", &a, &b)
 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 #define pint(a) printf("%d\n", a)
 #define test_print1(a) cout << "var1 = " << a << endl
 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl

 typedef long long LL;
 typedef pair<int, int> pii;
 typedef vector<int> vi;

 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 3e4 + ;
 const int md = ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 int make_id(int x, int y, int n) { return x * n + y; }

 LL ans;

 struct AhoCorasickAutoMata {
     const static int maxn = ;
     int cc;
     map<int, int> ch[maxn];
     vector<int> son[maxn];
     int val[maxn], last[maxn], f[maxn];

     void clear() { cc = ; mem0(val); mem0(ch); mem0(last); mem0(f); rep_up0(i, maxn) { son[i].clear(); ch[i].clear(); } }

     void Insert(int s[], int n) {
         int pos = ;
         rep_up0(i, n) {
             int id = s[i];
             if(!ch[pos][id]) {
                 ch[pos][id] = ++ cc;
                 son[pos].push_back(id);
             }
             pos = ch[pos][id];
         }
         val[pos] ++;
     }

     void print(int j) {
         if (j) {
             ans += val[j];
             print(last[j]);
         }
     }

     void find(int T[], int n) {
         int j = ;
         rep_up0(i, n) {
             int c = T[i];
             while (j && !ch[j][c]) j = f[j];
             j = ch[j][c];
             if (val[j]) print(j);
             else {
                 if (last[j]) print(last[j]);
             }
         }
     }

     int getFail() {
         queue<int> q;
         f[] = ;
         int SZ = son[].size();
         rep_up0(i, SZ) {
             int c = son[][i];
             int u = ch[][c];
             q.push(u);
         }
         while (!q.empty()) {
             int r = q.front(), SZ = son[r].size(); q.pop();
             rep_up0(i, SZ) {
                 int c = son[r][i];
                 int u = ch[r][c];
                 q.push(u);
                 int v = f[r];
                 while (v && !ch[v][c]) v = f[v];
                 f[u] = ch[v][c];
                 last[u] = val[f[u]]? f[u] : last[f[u]];
             }
         }
     }
 };

 AhoCorasickAutoMata ac;
 map<pii, int> mp;
 int pa[], pb[];
 int a[], b[];

 int main() {
     //freopen("in.txt", "r", stdin);
     int T, n, m, tot = ;
     cin >> T;
     while (T--) {
         ans = ;
         ac.clear();
         mp.clear();
         cin >> n >> m;
         rep_up0(i, n) {
             sint(a[i]);
         }
         rep_up0(i, n - ) {
             int g = gcd(a[i], a[i + ]);
             int ga = a[i] / g, gb = a[i + ] / g;
             int &x = mp[make_pair(ga, gb)];
             if (!x) x = ++ tot;
             pa[i] = x;
         }

         rep_up0(i, m) {
             int k;
             sint(k);
             rep_up0(j, k) {
                 sint(b[j]);
             }
             if (k > n) continue;
             if (k == ) {
                 ans += n;
                 continue;
             }
             rep_up0(j, k - ) {
                 int g = gcd(b[j], b[j + ]);
                 int ga = b[j] / g, gb = b[j + ] / g;
                 int &x = mp[make_pair(ga, gb)];
                 if (!x) x = ++ tot;
                 pb[j] = x;
             }
             ac.Insert(pb, k - );
         }
         ac.getFail();
         ac.find(pa, n - );
         cout << ans << endl;
     }
     return ;
 }