ZOJ3784 String of Infinity(AC自动机&&强连通分量)

题意：给你n个禁止串，然后你只能用字符表的前m个字符去写一个无限长的串，要求是不能包含禁止串，而且串在后面不能出现循环

比赛的时候想的是先建一个自动机，然后将自动机确定化，不能到达的状态全部弄出来。但是对于剩下的状态就卡住了，我怎么才能知道这些状态会构成循环呢？后来看了别人的代码，看到了强连通分量，我就恍然大悟了。其实只需要对剩下的未确定的状态，根据转移边建图，然后跑一次强连通分量。

这么做的效果就是将原图剩下的状态缩成了一个DAG，我们每次只能由根结点往下走，我们必然需要停留在某个强连通分量里，不然的话我走到拓扑序最后的分量就不能再走了（或者说走到拓扑序最后的话，就只能在那个强连通分量沿着分量内的边走“自环”）。如果这个强连通分量是一个环的话，那么显然我们停留在这个强连通分量里的话就会有循环节。所以问题转化为是否存在一个强连通分量它不是一个环。判断方法就是维系一个大小为n的强连通分量至少需要n条边，只要强连通分量内的边大于n，它就不是一个自环。

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#pragma warning(disable:4996) #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <set> #include <vector> #include<queue> #include<stack> #include <cmath> using namespace std;   #define maxn 210000 #define ll long long   int n, m;   struct Trie{     Trie *fail, *go[26];     bool ter; bool flag;     void init(){         memset(go, 0, sizeof(go)); fail = NULL; ter = false; flag = false;     } }pool[maxn], *root; int tot;   void insert(char *c){     int len = strlen(c); Trie *p = root;     for (int i = 0; i < len; i++){         if (p->go[c[i] - 'a'] != 0) p = p->go[c[i] - 'a'];         else{             pool[tot].init();             p->go[c[i] - 'a'] = &pool[tot++];             p = p->go[c[i] - 'a'];         }     }     p->ter = true; }   void getFail() {     queue<Trie*> que;     que.push(root);     root->fail = NULL;     while (!que.empty()){         Trie *temp = que.front(); que.pop();         Trie *p = NULL;         for (int i = 0; i < m; i++){             if (temp->go[i] != NULL){                 if (temp == root) temp->go[i]->fail = root;                 else{                     p = temp->fail;                     while (p != NULL){                         if (p->go[i] != NULL){                             temp->go[i]->fail = p->go[i]; break;                         }                         p = p->fail;                     }                     if (p == NULL) temp->go[i]->fail = root;                 }                 que.push(temp->go[i]);             }         }     } }   bool ddfs(Trie *p){     if (p == root||p==NULL) return false;     if (p->flag == true) return p->ter;     p->ter |= ddfs(p->fail); p->flag = true;     return p->ter; }   int pre[maxn], low[maxn], sccno[maxn],siz[maxn]; int sta[maxn],st; int dfs_clock; int scc_cnt;   int siz2[maxn];   void dfs(int u){     low[u] = pre[u] = ++dfs_clock;     sta[++st] = u;     for (int i = 0; i < m; i++){         Trie *p = pool[u].go[i];         if (p->ter) continue;         int v = p - pool;         if (!pre[v]){             dfs(v); low[u] = min(low[u], low[v]);         }         else if (!sccno[v]){             low[u] = min(low[u], pre[v]);         }     }     if (low[u] == pre[u]){         ++scc_cnt;         while (1){             int x = sta[st--]; sccno[x] = scc_cnt;             siz[scc_cnt]++;             if (x == u) break;         }     } } void find_scc() {     memset(siz, 0, sizeof(siz));     memset(sccno, 0, sizeof(sccno));     memset(low, 0, sizeof(low));     memset(pre, 0, sizeof(pre));     st = dfs_clock = 0; scc_cnt = 0;     for (int i = 0; i < tot; i++){         if (pool[i].ter) continue;         if (!pre[i]) dfs(i);     } }   char str[1500];   int main() {     int T; cin >> T;     while (T--)     {         cin >> n >> m;         tot = 0; root = &pool[tot++]; root->init();         for (int i = 0; i < n; i++){             scanf("%s", str);             insert(str);         }         getFail();         for (int i = 0; i < tot; i++) ddfs(&pool[i]);         for (int i = 0; i < tot; i++){             Trie *p = &pool[i];             for (int k = 0; k < m; k++){                 if (p->go[k] == NULL){                     Trie *temp = p; temp = temp->fail;                     while (temp != NULL){                         if (temp->go[k] != NULL) {                             p->go[k] = temp->go[k]; break;                         }                         temp = temp->fail;                     }                     if (temp == NULL) p->go[k] = root;                 }             }         }         find_scc();         memset(siz2, 0, sizeof(siz2));         for (int i = 0; i < tot; i++){             if (pool[i].ter) continue;             for (int j = 0; j < m; j++){                 int k = pool[i].go[j] - pool;                 if (sccno[k] == sccno[i]){                     siz2[sccno[k]]++;                 }             }         }         bool flag = false;         for (int i = 1; i <= scc_cnt; i++){             if (siz2[i]>siz[i]){                 flag = true; break;             }         }         if (flag) puts("Yes");         else puts("No");     }     return 0; }