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  • P2444 [POI2000]病毒(AC自动机)

    https://www.luogu.com.cn/problem/P2444

    直接AC自动机即可。
    建完后,跑一边AC机,看一下能不能跑出一个不经过病毒,并且循环的字符串。

     1 #include <bits/stdc++.h>
     2 
     3 using namespace std;
     4 const int maxn = 3e4 + 5;
     5 int n;
     6 char s[maxn];
     7 int cnt;
     8 struct node {
     9     int fail, flag, son[2] = {-1,-1};
    10 } tree[maxn];
    11 
    12 queue<int> q;
    13 int hh[maxn];
    14 bool v[maxn];
    15 
    16 void insert(char *s) {
    17     int u = 0, len = strlen(s + 1);
    18     for (int i = 1; i <= len; i++) {
    19         int  v = s[i] - '0';
    20         if (tree[u].son[v] == -1)
    21             tree[u].son[v] = ++cnt;
    22         u = tree[u].son[v];
    23     }
    24     tree[u].flag = 1;
    25 }
    26 
    27 void getfail() {
    28     q.push(0);
    29     while (!q.empty()) {
    30         int u = q.front();
    31         q.pop();
    32         for (int i = 0; i < 2; i++) {
    33             int v = tree[u].son[i];
    34             if (v == -1) continue;
    35             if (!u) tree[v].fail = 0;
    36             else {
    37                 int j = tree[u].fail;
    38                 while (j && tree[j].son[i] == -1)
    39                     j = tree[j].fail;
    40                 tree[v].fail = max(tree[j].son[i], 0);
    41             }
    42             q.push(v);
    43         }
    44         if (tree[tree[u].fail].flag == 1) tree[u].flag = 1;
    45     }
    46 }
    47 
    48 bool dfs(int x) {
    49     v[x] = 1;
    50     for (int i = 0; i < 2; i++) {
    51         int j = x;
    52         while (j && tree[j].son[i] == -1) j = tree[j].fail;
    53         j = tree[j].son[i];
    54         if (j == -1) j = 0;
    55         if (v[j]) return 1;
    56         if (hh[j] || tree[j].flag == 1) continue;
    57         hh[j] = 1;
    58         if (dfs(j)) return 1;
    59     }
    60     v[x] = 0;
    61     return 0;
    62 }
    63 
    64 int main() {
    65     //freopen("in","r",stdin);
    66     scanf("%d",&n);
    67     for (int i = 1; i <= n; i++) {
    68         scanf("%s", s + 1);
    69         insert(s);
    70     }
    71     getfail();
    72     if (dfs(0)) puts("TAK");
    73     else puts("NIE");
    74     return 0;
    75 }
    View Code
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  • 原文地址:https://www.cnblogs.com/xcfxcf/p/12534648.html
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