标签:增加 english hid 建立 ever review 返回 position code
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1402 Accepted Submission(s): 754
#include <stdio.h> #include <algorithm> #include <iostream> #include <string.h> #include <queue> using namespace std; typedef long long ll; struct Trie { int Next[500010][26];//26是这里讨论26个小写字母的情况,根据情况修改 int fail[500010],end[500010];//end数组表示以该节点结尾的字符串的数量 int root,L;//L用来标记节点序号,以广度优先展开的字典树的序号 int newnode() //建立新节点 { for(int i = 0;i < 26;i++) Next[L][i] = -1; //将该节点的后继节点域初始化 end[L++] = 0; return L-1; //返回当前节点编号 } void init() //初始化操作 { L = 0; root = newnode(); } void insert(char buf[]) { int len = strlen(buf); int now = root; for(int i = 0;i < len;i++) { if(Next[now][buf[i]-‘a‘] == -1) //如果未建立当前的后继节点,建立新的节点 Next[now][buf[i]-‘a‘] = newnode(); now = Next[now][buf[i]-‘a‘]; } end[now]++;//以该节点结尾的字符串数量增加1 } void build() { queue<int>Q; //用广度优先的方式,将树层层展开 fail[root] = root; for(int i = 0;i < 26;i++) 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(int i = 0;i < 26;i++) if(Next[now][i] == -1) Next[now][i] = Next[fail[now]][i];//该段的最后一个节点匹配后,跳到拥有最大公共后缀的fail节点继续匹配 else { fail[Next[now][i]]=Next[fail[now]][i];//当前节点的fail节点等于它前驱节点的fail节点的后继节点 Q.push(Next[now][i]); } } } ll query(string buf) { int len = buf.size(); int now = root; ll res = 0; for(int i = 0;i < len;i++) { now = Next[now][buf[i]-‘a‘]; int temp = now; while( temp != root ) { res += end[temp];//加上以当前节点结尾的字符串数 temp = fail[temp];//每次找最大公共后缀对应的fail节点 } } return res; } void debug() { for(int i = 0;i < L;i++) { printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]); for(int j = 0;j < 26;j++) printf("%2d",Next[i][j]); printf("]\n"); } } }; string buf[100010]; Trie ac; char str[100010]; int main() { int t,n,ans,m; scanf("%d",&t); while(t--) { ac.init(); scanf("%d%d",&n,&m); for(int i=0;i<n;i++) cin>>buf[i]; for(int i=0;i<m;i++){ scanf("%s",str); ac.insert(str); } ac.build(); for(int i=0;i<n;i++) { cout<<ac.query(buf[i])<<endl; } } return 0; }
标签:增加 english hid 建立 ever review 返回 position code
原文地址:http://www.cnblogs.com/a249189046/p/7623349.html