zoj3228Searching the String（ac自动机)

链接

这个题把病毒分为了两种，一种包含可以覆盖，另一种不可以，需要分别求出包含他们的个数，可以把两种都建在一颗tire树上，在最后求得时候判断一下当前节点是属于哪种字符串，如果是不包含的需要判断一下pre[i]+len[i]<=当前位置。

注意会有重复字符串，可以先map处理一下。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 #include<string>
 #include<map>
 using namespace std;
 #define N 600110
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 const int child_num = ;
 char vir[];
 char s[N/];
 int o[N/],len[N/],ans[N/][],po[N/];
 int cc[N/];
 map<string,int>f;
 vector<int>ed[N/];
 class AC
 {
     private:
     int ch[N][child_num];
     int fail[N];
     int Q[N];
     int val[N];
     int sz;
     int id[];
     public:
     void init()
     {
         fail[] = ;
         for(int i =  ;i < child_num ; i++)
         id[i+'a'] = i;
     }
     void reset()
     {
         memset(ch[],,sizeof(ch[]));
         memset(val,,sizeof(val));
         sz = ;
     }
     void insert(char *a,int key)
     {
         int p = ;
         for(; *a ; a++)
         {
             int d= id[*a];
             if(ch[p][d]==)
             {
                 memset(ch[sz],,sizeof(ch[sz]));
                 ch[p][d] = sz++;
             }
             p = ch[p][d];
         }
         val[p] = key;
     }
     void construct()
     {
         int i,head=,tail = ;
         for(i = ; i  < child_num ;i++)
         {
             if(ch[][i])
             {
                 fail[ch[][i]] = ;
                 Q[tail++] = ch[][i];
             }
         }
         while(head!=tail)
         {
             int u = Q[head++];
             for(i = ; i < child_num ; i++)
             {
                 if(ch[u][i])
                 {
                     fail[ch[u][i]] = ch[fail[u]][i];
                     Q[tail++] = ch[u][i];
                 }
                 else ch[u][i] = ch[fail[u]][i];
             }
         }
     }
     void work(int m,int kk,int g)
     {
         memset(ans,,sizeof(ans));
         memset(po,-,sizeof(po));
         int p = ,i,k = strlen(s);
         for(i =  ; i < k ; i++)
         {
             int d = id[s[i]];
             p = ch[p][d];
             int tmp = p;
             while(tmp)
             {
                 int v = val[tmp];
                 ans[v][]++;
                 if(po[v]+len[v]<=i)
                 {
                     po[v] = i;
                     ans[v][]++;
                 }
                 tmp = fail[tmp];
             }
         }
         for(i = ; i <= g ; i++)
         {
             for(int j = ; j < ed[i].size() ; j++)
             {
                 int v = ed[i][j];
                 if(o[v]) cc[v] = ans[i][];
                 else cc[v] = ans[i][];
             }
         }
         printf("Case %d\n",kk);
         for(i =  ; i <= m ;i++)
         printf("%d\n",cc[i]);
         puts("");
     }
 }ac;
 int main()
 {
     int i,m,kk=;
     ac.init();
     while(scanf("%s",s)!=EOF)
     {
         ac.reset();
         f.clear();
         scanf("%d",&m);
         int g = ;
         for(i =  ; i <= m; i++)
         ed[i].clear();
         for(i = ;i <= m ;i++)
         {
             scanf("%d%s",&o[i],vir);
             if(!f[vir])
             {
                 f[vir] = (++g);
                 ac.insert(vir,g);
                 len[g] = strlen(vir);
             }
             ed[f[vir]].push_back(i);
         }
         ac.construct();
         kk++;
         ac.work(m,kk,g);
     }
     return ;
 }