HDU 2222 Keywords Search(瞎搞)
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Keywords Search
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 50451 Accepted Submission(s): 16236
Problem Description
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.
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.
Input
First 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.
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.
Output
Print how many keywords are contained in the description.
Sample Input
1
5
she
he
say
shr
her
yasherhs
Sample Output
3
Author
Wiskey
/* *********************************************** Author :guanjun Created Time :2016/5/26 22:17:48 File Name :2222.cpp ************************************************ */ #include <iostream> #include <cstring> #include <cstdlib> #include <stdio.h> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <iomanip> #include <list> #include <deque> #include <stack> #define ull unsigned long long #define ll long ling #define mod 90001 #define INF 0x3f3f3f3f #define maxn 250010 #define cle(a) memset(a,0,sizeof(a)) const ull inf = 1LL << 61; const double eps=1e-5; using namespace std; struct ACauto{ int ch[maxn][26]; int sz; int f[maxn],last[maxn],val[maxn],cnt[maxn]; void init(){ sz=1; memset(ch[0],0,sizeof ch[0]); memset(cnt,0,sizeof cnt); } int idx(char c){ return c-‘a‘; } void add(char *s,int v){ int u=0,len=strlen(s); for(int i=0;i<len;i++){ int c=idx(s[i]); if(!ch[u][c]){ memset(ch[sz],0,sizeof ch[sz]); val[sz]=0; ch[u][c]=sz++; } u=ch[u][c]; } val[u]=v; } void getfail(){ queue<int>q; f[0]=0; for(int c=0;c<26;c++){ int u=ch[0][c]; if(u){ f[u]=0; q.push(u); last[u]=0; } } while(!q.empty()){ int r=q.front();q.pop(); for(int c=0;c<26;c++){ int u=ch[r][c]; if(!u){ ch[r][c]=ch[f[r]][c]; continue; } q.push(u); f[u]=ch[f[r]][c]; last[u]=val[f[u]]?f[u]:last[f[u]]; } } } void print(int j){ if(j){ cnt[val[j]]++; print(last[j]); } } void Find(char *T){ int n=strlen(T); int j=0; for(int i=0;i<n;i++){ int c=idx(T[i]); while(j&&!ch[j][c])j=f[j]; j=ch[j][c]; if(val[j])print(j); else if(last[j])print(last[j]); } } }ac; char s[10010][60],T[1000010]; int BKDRHash(char* s) { long long seed=131; long long hash=0; while(*s==‘0‘)s++; while(*s) { hash=hash*seed+(*s++); } return (hash & 0x7FFFFFFF); } map<int ,int>mp; int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif //freopen("out.txt","w",stdout); int t,n; cin>>t; while(t--){ ac.init(); mp.clear(); scanf("%d",&n); for(int i=1;i<=n;i++){ scanf("%s",s[i]); ac.add(s[i],i); int x=BKDRHash(s[i]); mp[x]++; } ac.getfail(); scanf("%s",T); ac.Find(T); int ans=0; for(int i=1;i<=n;i++){ int x=ac.cnt[i]; int y=BKDRHash(s[i]); if(x>0){ ans+=mp[y]; } } printf("%d\n",ans); } return 0; }
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