Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 60242    Accepted Submission(s): 19853 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.   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.   Output Print how many keywords are contained in the description.   Sample Input 1 5 she he say shr her yasherhs   Sample Output 3   题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=2222 题意：一个字符串和n个匹配串，问最多有几个可以匹配的。 思路：ac自动机，网上有很详细的介绍，我看的是： http://blog.csdn.net/mobius_strip/article/details/22549517 这题只是用来熟悉下模板的。 小问题：               数组实现线段树时不应该初始全部赋值为0,否则会超时。 代码： #include<bits/stdc++.h> using namespace std; const int maxn=1000000+100; int tree[maxn][30],v[maxn],fail[maxn],last[maxn]; int cnt,ans; char s[maxn]; void init() { cnt=0; ans=0; memset(tree[0],0,sizeof(tree[0])); memset(v,0,sizeof(v)); memset(last,0,sizeof(last)); memset(fail,0,sizeof(fail)); } void build(char *T) { int n=strlen(T); int j=0; for(int i=0; i<n; i++) { int k=tree[j][T[i]-'a']; if(!k) { tree[j][T[i]-'a']=++cnt; memset(tree[cnt],0,sizeof(tree[cnt])); } j=tree[j][T[i]-'a']; } v[j]++; } void getfail() { queue<int>q; for(int i=0; i<26; i++) { if(tree[0][i]) q.push(tree[0][i]); } while(!q.empty()) { int u=q.front(); q.pop(); for(int i=0; i<26; i++) { int vv=tree[u][i]; if(!vv) { tree[u][i]=tree[fail[u]][i]; continue; } q.push(vv); int uu=fail[u]; while(uu&&!tree[uu][i])uu=fail[uu]; fail[vv]=tree[uu][i]; last[vv]=v[fail[vv]]?fail[vv]:last[fail[vv]]; //找到一个之后继续匹配相同前缀的项 } } } void findf(int j) { if(!j)return ; if(v[j]) { ans+=v[j]; v[j]=0; } findf(last[j]); } void solve(char* T) { getfail(); int n=strlen(T),j=0; for(int i=0;i<n;i++) { j=tree[j][T[i]-'a']; if(v[j])findf(j); else if(last[j])findf(last[j]); } } int main() { int T; scanf("%d",&T); while(T--) { init(); int n; scanf("%d",&n); for(int i=0; i<n; i++) { scanf("%s",s); build(s); } scanf("%s",s); solve(s); printf("%d\n",ans); } return 0; }