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Problem DescriptionIn the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.
InputFirst line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters ‘a‘-‘z‘, and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.
OutputPrint how many keywords are contained in the description.
Sample Input15shehesayshrheryasherhs
Sample Output3
4 5 she he say shr her yasherhs 5 she he say shr her yasherhs 3 she she she shesheshe 6 she he he say shr her yasherhs 答案: 3 3 3 4
代码如下:
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<iostream> 5 #include<algorithm> 6 #include<queue> 7 using namespace std; 8 #define Maxn 800100 9 #define Maxl 60 10 11 char s[Maxl]; 12 char ss[1000010]; 13 14 struct node 15 { 16 int son[30],cnt,fail; 17 }t[Maxn];int tot; 18 19 void upd(int x) 20 { 21 t[x].cnt=0; 22 memset(t[x].son,0,sizeof(t[x].son)); 23 } 24 25 void read_trie() 26 { 27 int n; 28 scanf("%d",&n); 29 tot=0;upd(0); 30 for(int i=1;i<=n;i++) 31 { 32 int now; 33 scanf("%s",s+1); 34 int len=strlen(s+1); 35 now=0; 36 for(int j=1;j<=len;j++) 37 { 38 int ind=s[j]-‘a‘+1; 39 if(!t[now].son[ind]) 40 t[now].son[ind]=++tot,upd(tot); 41 now=t[now].son[ind]; 42 if(j==len) t[now].cnt++; 43 } 44 } 45 } 46 47 queue<int > q; 48 void build_AC() 49 { 50 int i,j,x,y; 51 while(!q.empty()) q.pop(); 52 q.push(0); 53 while(!q.empty()) 54 { 55 x=q.front(); 56 y=t[x].fail; 57 for(j=1;j<=26;j++) 58 { 59 if(t[x].son[j]) 60 { 61 q.push(t[x].son[j]); 62 t[t[x].son[j]].fail=x?t[y].son[j]:x; 63 } 64 else t[x].son[j]=t[y].son[j]; 65 } 66 q.pop(); 67 } 68 } 69 70 int gmark(int x,int y) 71 { 72 if(!x) return y; 73 if(t[x].cnt) y+=t[x].cnt,t[x].cnt=0; 74 return gmark(t[x].fail,y); 75 } 76 77 void ffind() 78 { 79 scanf("%s",ss+1); 80 int l=strlen(ss+1); 81 int now,ans=0; 82 for(int i=1;i<=l;i++) 83 { 84 int x=ss[i]-‘a‘+1; 85 now=t[now].son[x]; 86 ans+=gmark(now,0); 87 } 88 printf("%d\n",ans); 89 } 90 91 int main() 92 { 93 int T; 94 scanf("%d",&T); 95 while(T--) 96 { 97 read_trie(); 98 build_AC(); 99 ffind(); 100 } 101 return 0; 102 }
2016-06-14 17:02:09
【HDU2222】Keywords Search(AC自动机)
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原文地址:http://www.cnblogs.com/Konjakmoyu/p/5584658.html