大二训练第一周 G - Keywords Search 艾斯atman
2015-11-07 14:08
423 查看
G - Keywords Search
Time Limit:1000MS Memory Limit:131072KB 64bit IO Format:%I64d
& %I64u
Submit Status
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
AC奥特曼 tire+kmp
ACcode:
换个清晰的模板
Time Limit:1000MS Memory Limit:131072KB 64bit IO Format:%I64d
& %I64u
Submit Status
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
AC奥特曼 tire+kmp
ACcode:
#pragma warning(disable:4786)//使命名长度不受限制 #pragma comment(linker, "/STACK:102400000,102400000")//手工开栈 #include <map> #include <set> #include <queue> #include <cmath> #include <stack> #include <cctype> #include <cstdio> #include <cstring> #include <stdlib.h> #include <iostream> #include <algorithm> #define rd(x) scanf("%d",&x) #define rd2(x,y) scanf("%d%d",&x,&y) #define rds(x) scanf("%s",x) #define rdc(x) scanf("%c",&x) #define ll long long int #define maxn 500015 #define mod 1000000007 #define INF 0x3f3f3f3f //int 最大值 #define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i) #define MT(x,i) memset(x,i,sizeof(x)) #define PI acos(-1.0) #define E exp(1) #define MAX 26 using namespace std; struct Trie{ int next[maxn][MAX],fail[maxn],end[maxn]; int root,L; int newnode(){ FOR(i,0,MAX-1)next[L][i]=-1; end[L++]=0; return L-1; } void init(){ L=0; root=newnode(); } void insert(char *buf){ int now=root; for(int i=0;buf[i]!='\0';++i){ if(next[now][buf[i]-'a']==-1) next[now][buf[i]-'a']=newnode(); now=next[now][buf[i]-'a']; } end[now]++; } void build(){ queue<int> Q; fail[root]=root; FOR(i,0,MAX-1) if(next[root][i]==-1) next[root][i]=root; else { fail[next[root][i]]=root; Q.push(next[root][i]); } while(!Q.empty()){ int now=Q.front(); Q.pop(); FOR(i,0,MAX-1) if(next[now][i]==-1) next[now][i]=next[fail[now]][i]; else { fail[next[now][i]]=next[fail[now]][i]; Q.push(next[now][i]); } } } int query(char *buf){ int len=strlen(buf); int now=root; int res=0; FOR(i,0,len-1){ now=next[now][buf[i]-'a']; int temp=now; while(temp!=root){ res+=end[temp]; end[temp]=0; temp=fail[temp]; } } return res; } }; char str[1000000+100]; Trie AC_Automan; int main(){ int loop,n;rd(loop); while(loop--){ rd(n);AC_Automan.init(); FOR(i,1,n){ rds(str); AC_Automan.insert(str); } AC_Automan.build();rds(str); printf("%d\n",AC_Automan.query(str)); } return 0; } /* 1 5 she he say shr her yasherhs */
换个清晰的模板
#include <iostream> #include <cstring> #include <queue> #include <cstdio> #define ACnodeQueue queue<int> #define MAX 26 #define maxn 1000100 using namespace std; struct ACnode{ ACnode *fail; ACnode *next[MAX]; int id, val, cnt; void reset(int _id){ id=_id; cnt=val=0; fail=NULL; memset(next,0,sizeof(next)); } bool isReceiving(){return val;} }; class ACAutoman{ public: ACnode *nodes[maxn]; int nodesMax; ACnode *root; int node_cnt; int ID[256],IDsize; public: ACAutoman(){nodesMax=0;} void init(){ node_cnt=IDsize=0; root=getNode(); memset(ID,-1,sizeof(ID)); } ACnode *getNode(){ if(node_cnt>=nodesMax){ nodes[nodesMax++]=new ACnode(); } ACnode *p=nodes[node_cnt]; p->reset(node_cnt++); return p; } int getCharID(unsigned char c,int needcreate){ if(!needcreate)return ID[c]; return ID[c]!=-1?ID[c]:ID[c]=IDsize++; } void insert(char *str,int val){ ACnode *p=root; int id; for(int i=0;str[i];++i){ id=getCharID(str[i],true); if(p->next[id]==NULL)p->next[id]=getNode(); p=p->next[id]; } p->val|=val; p->cnt++; } void bfs(){ queue<ACnode *>q; ACnode *now,*son,*tmp; root->fail=NULL; q.push(root); while(!q.empty()){ now=q.front(); q.pop(); if(now->fail){ if(now->isReceiving()){ for(int i=0;i<MAX;++i) now->next[i]=now; continue; } } for(int i=0;i<MAX;++i){ son=now->next[i]; tmp=(now==root)?root:now->fail->next[i]; if(son==NULL)now->next[i]=tmp; else { son->fail=tmp; q.push(son); } } } } int query_str(char *str){ ACnode *p=root,*tmp=NULL; int id; int cnt=0; for(int i=0;str[i];++i){ id=getCharID(str[i],false); if(id==-1){ p=root; continue; } p=p->next[id]; tmp=p; while(tmp!=root&&tmp->cnt!=-1){ if(tmp->cnt){ cnt+=tmp->cnt; tmp->cnt=-1; } tmp=tmp->fail; } } return cnt; } }AC; char t[maxn]; int main(){ int loop; scanf("%d",&loop); while(loop--){ int n; scanf("%d",&n); AC.init(); char tmp[55]; for(int i=0;i<n;++i){ scanf("%s",tmp); AC.insert(tmp,0); } AC.bfs(); scanf("%s",t); printf("%d\n",AC.query_str(t)); } return 0; }
相关文章推荐
- 正则判断手机号
- 阿里云免费服务器搭建学习过程--成功:
- android-ViewPager实现图片自动切换
- mysql学习笔记(1)
- GTK+图形化应用程序开发学习笔记(十一)—箭头构件、工具提示对象
- Java记录 -63- Java的键值映射Map
- php实现点击可刷新验证码
- iOS调试技巧之打印输出 -----A: (NSString *)description B:自定义LOG C:使用第三方插件快速打印
- 封装
- GTK+图形化应用程序开发学习笔记(十)—工具条
- [C++ Mind Map] class and memory
- 学习笔记
- 如何避免上传gif到数据库时变成静态图片
- HyperlinkButton——WP8控件学习
- mac安装brew
- 当在Win8下安装msi类型的文件出现errorcode 2503的解决方法
- va_start和va_end使用详解
- Daily Scrum (2015/11/6)
- python 使用多线程进行压力测试
- @RequestParam @RequestBody @PathVariable 等参数绑定注解详解(转)