xtu字符串 D. 病毒侵袭

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cstdlib>
  5 #include <vector>
  6 #include <climits>
  7 #include <algorithm>
  8 #include <cmath>
  9 #include <queue>
 10 #define LL long long
 11 #define INF 0x3f3f3f
 12 using namespace std;
 13 const int maxn = 100000;
 14 struct trie {
 15     int cnt,id,wd[130],fail;
 16     void init() {
 17         id = cnt = 0;
 18         fail = -1;
 19         memset(wd,-1,sizeof(wd));
 20     }
 21 } dic[maxn];
 22 int tot,ans[1100],total;
 23 void insertWord(int root,int _id,char *s) {
 24     for(int i = 0; s[i]; i++) {
 25         int k = s[i] - 31;
 26         if(dic[root].wd[k] == -1) {
 27             dic[tot].init();
 28             dic[root].wd[k] = tot++;
 29         }
 30         root = dic[root].wd[k];
 31     }
 32     dic[root].cnt++;
 33     dic[root].id = _id;
 34 }
 35 void build(int root) {
 36     queue<int>q;
 37     q.push(root);
 38     while(!q.empty()) {
 39         int u = q.front();
 40         q.pop();
 41         for(int i = 0; i < 130; i++) {
 42             if(dic[u].wd[i] == -1) continue;
 43             if(!u) dic[dic[u].wd[i]].fail = 0;//如果是第二层的节点
 44             else {
 45                 int v = dic[u].fail;
 46                 while(v && dic[v].wd[i] == -1)
 47                     v = dic[v].fail;
 48                 //回溯到离根较远并与当前字符相同的点
 49                 if(dic[v].wd[i] != -1)
 50                     dic[dic[u].wd[i]].fail = dic[v].wd[i];
 51                 else dic[dic[u].wd[i]].fail = 0;
 52             }
 53             q.push(dic[u].wd[i]);
 54         }
 55     }
 56 }
 57 
 58 void query(int root,char *s) {
 59     bool vis[510] = {false};
 60     for(int i = 0; s[i]; i++) {
 61         int k = s[i] - 31;
 62         while(root && dic[root].wd[k] == -1)
 63             root = dic[root].fail;//不如当前字符匹配，回溯
 64         root = dic[root].wd[k];//dic[root].wd[k]与当前字符匹配
 65         if(root == -1) root = 0;//trie树上不存在与之匹配的
 66         else {
 67             int v = root;
 68             while(v && !vis[dic[v].id]) {
 69                 //如果当前节点访问过了，
 70                 //从当前节点的回溯路径上的节点也被访问了
 71                 if(dic[v].cnt) {
 72                     vis[dic[v].id] = true;
 73                     ans[total++] = dic[v].id;
 74                 }
 75                 v = dic[v].fail;
 76             }
 77         }
 78     }
 79 }
 80 int main() {
 81     int n,m,i,j,t = 0;
 82     char word[300],text[11000];
 83     scanf("%d",&n);
 84     dic[0].init();
 85     tot = 1;
 86     for(i = 1; i <= n; i++) {
 87         scanf("%s",word);
 88         insertWord(0,i,word);
 89     }
 90     build(0);
 91     scanf("%d",&m);
 92     for(i = 1; i <= m; i++) {
 93         total = 0;
 94         scanf("%s",text);
 95         query(0,text);
 96         if(total) {
 97             t++;
 98             sort(ans,ans+total);
 99             printf("web %d:",i);
100             for(j = 0; j < total; j++)
101                 printf(" %d",ans[j]);
102             printf("\n");
103         }
104     }
105     printf("total: %d\n",t);
106     return 0;
107 }