标签:比较 with ace integer eof push 数组实现 one std
In 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 Input
1 5 she he say shr her yasherhs
Sample Output
3
AC自动机裸题,下面给出用指针实现和用数组实现两种方法。指针比较容易理解,但数组不会出现各种奇奇怪怪的玄学问题。
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <queue> 5 #include <algorithm> 6 7 using namespace std; 8 char str[1000000+100]; 9 10 struct node 11 { 12 int num; 13 struct node *next[26]; 14 struct node *fail; 15 void init() 16 { 17 int i; 18 for(i=0;i<26;i++) 19 next[i]=NULL; 20 num=0; 21 fail=NULL; 22 } 23 }*root; 24 25 26 node* rt; 27 int id,len; 28 29 void build() 30 { 31 rt=root; 32 len=strlen(str); 33 for(int i=0;i<len;i++) 34 { 35 id=str[i]-‘a‘; 36 if(rt->next[id]==NULL) 37 { 38 rt->next[id]=new node(); 39 rt->next[id]->init(); 40 } 41 rt=rt->next[id]; 42 } 43 rt->num++; 44 } 45 46 void get_fail() 47 { 48 rt=root; 49 node *son,*temp; 50 queue<node *> q; 51 q.push(rt); 52 while(!q.empty()) 53 { 54 temp=q.front(); 55 q.pop(); 56 for(int i=0;i<26;i++) 57 { 58 son=temp->next[i]; 59 if(son!=NULL) 60 { 61 if(temp==root) 62 son->fail=root; 63 else 64 { 65 rt=temp->fail; 66 while(rt) 67 { 68 if(rt->next[i]) 69 { 70 son->fail=rt->next[i]; 71 break; 72 } 73 rt=rt->fail; 74 } 75 if(!rt) 76 son->fail=root; 77 } 78 q.push(son); 79 } 80 } 81 } 82 } 83 84 int querry() 85 { 86 rt=root; 87 len=strlen(str); 88 node *temp; 89 int ans=0; 90 for(int i=0;i<len;i++) 91 { 92 id=str[i]-‘a‘; 93 while(!rt->next[id]&&rt!=root) 94 rt=rt->fail; 95 rt=rt->next[id]; 96 if(!rt) 97 rt=root; 98 temp=rt; 99 while(temp!=root) 100 { 101 if(temp->num>=0) 102 { 103 ans+=temp->num; 104 temp->num=-1; 105 } 106 else 107 break; 108 temp=temp->fail; 109 } 110 } 111 return ans; 112 } 113 114 int main() 115 { 116 int T; 117 scanf("%d",&T); 118 while(T--) 119 { 120 int n; 121 root=new node; 122 root->init(); 123 scanf("%d",&n); 124 getchar(); 125 for(int i=0;i<n;i++) 126 { 127 gets(str); 128 build(); 129 } 130 get_fail(); 131 gets(str); 132 printf("%d\n",querry()); 133 } 134 return 0; 135 }
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<iostream> 5 #include<queue> 6 using namespace std; 7 8 const int N=10100,L=1000100; 9 char s[L]; 10 int num,n; 11 struct node{ 12 int son[30]; 13 int fail,cnt; 14 }a[N*60]; 15 queue<int> q; 16 17 void clear(int x) 18 { 19 a[x].cnt=0; 20 a[x].fail=0; 21 memset(a[x].son,0,sizeof(a[x].son)); 22 } 23 24 void trie(char *c) 25 { 26 int l=strlen(c); 27 int x=0; 28 for(int i=0;i<l;i++) 29 { 30 int t=c[i]-‘a‘+1; 31 if(!a[x].son[t]) 32 { 33 num++; 34 clear(num); 35 a[x].son[t]=num; 36 } 37 x=a[x].son[t]; 38 } 39 a[x].cnt++; 40 } 41 42 void buildAC() 43 { 44 while(!q.empty()) q.pop(); 45 for(int i=1;i<=26;i++) 46 if(a[0].son[i]) q.push(a[0].son[i]); 47 while(!q.empty()) 48 { 49 int x=q.front();q.pop(); 50 int fail=a[x].fail; 51 for(int i=1;i<=26;i++) 52 { 53 int y=a[x].son[i]; 54 if(y) 55 { 56 a[y].fail=a[fail].son[i]; 57 q.push(y); 58 } 59 else a[x].son[i]=a[fail].son[i]; 60 } 61 } 62 } 63 64 int find(char *c) 65 { 66 int l=strlen(c); 67 int x=0,ans=0; 68 for(int i=0;i<l;i++) 69 { 70 int t=c[i]-‘a‘+1; 71 while(x && !a[x].son[t]) x=a[x].fail; 72 x=a[x].son[t]; 73 int p=x; 74 while(p && a[p].cnt!=-1) 75 { 76 ans+=a[p].cnt; 77 a[p].cnt=-1; 78 p=a[p].fail; 79 } 80 } 81 return ans; 82 } 83 84 int main() 85 { 86 int T; 87 scanf("%d",&T); 88 while(T--) 89 { 90 scanf("%d",&n); 91 num=0; 92 clear(0); 93 for(int i=1;i<=n;i++) 94 { 95 scanf("%s",s); 96 trie(s); 97 } 98 buildAC(); 99 scanf("%s",s); 100 printf("%d\n",find(s)); 101 } 102 return 0; 103 }
标签:比较 with ace integer eof push 数组实现 one std
原文地址:http://www.cnblogs.com/xibeiw/p/7434415.html