字典树sdutoj 1500模板
来源:互联网 发布:天下三捏脸数据 编辑:程序博客网 时间:2024/03/28 22:05
Message Flood
Time Limit: 1500ms Memory limit: 65536K 有疑问?点这里^_^
题目描述
Well, how do you feel about mobile phone? Your answer would probably be something like that "It's so convenient and benefits people a lot". However, If you ask Merlin this question on the New Year's Eve, he will definitely answer "What a trouble! I have to keep my fingers moving on the phone the whole night, because I have so many greeting message to send!" Yes, Merlin has such a long name list of his friends, and he would like to send a greeting message to each of them. What's worse, Merlin has another long name list of senders that have sent message to him, and he doesn't want to send another message to bother them Merlin is so polite that he always replies each message he receives immediately). So, before he begins to send message, he needs to figure to how many friends are left to be sent. Please write a program to help him. Here is something that you should note. First, Merlin's friend list is not ordered, and each name is alphabetic strings and case insensitive. These names are guaranteed to be not duplicated. Second, some senders may send more than one message to Merlin, therefore the sender list may be duplicated. Third, Merlin is known by so many people, that's why some message senders are even not included in his friend list.
输入
There are multiple test cases. In each case, at the first line there are two numbers n and m (1<=n,m<=20000), which is the number of friends and the number of messages he has received. And then there are n lines of alphabetic strings(the length of each will be less than 10), indicating the names of Merlin's friends, one per line. After that there are m lines of alphabetic strings, which are the names of message senders. The input is terminated by n=0.
输出
For each case, print one integer in one line which indicates the number of left friends he must send.
示例输入
5 3InkfishHenryCarpMaxJerichoCarpMaxCarp0
示例输出
3
#include <stdio.h>#include <stdlib.h>#include <string.h>struct code\\定义结构体{ int len; struct code *next[26];};struct code * newnode()\\节点初始化{ int i; struct code *p=(struct code *)malloc(sizeof(struct code)); p->len=0; for(i=0; i<26; i++) p->next[i]=NULL; return p;}void insert(struct code *root,char *s)\\插入并建树{ struct code *p=root; int i,len=strlen(s),t; for(i=0; i<len; i++) { if(s[i]>='a'&&s[i]<='z') t=s[i]-'a'; else t=s[i]-'A'; if(p->next[t]==NULL) p->next[t]=newnode(); p=p->next[t]; } p->len=1;}int find(struct code *root,char*s)\\查找{ struct code *p=root; int i,t,len=strlen(s); for(i=0; i<len; i++) { if(s[i]>='a'&&s[i]<='z') t=s[i]-'a'; else t=s[i]-'A'; if(p->next[t]==NULL) return 0; p=p->next[t]; } if(p->len) { len=0; return 1; } return 0;}int dealtrie(struct code *root)\\删除树,防止内存占用过多{ int i; if(root==NULL) return 0; for(i=0; i<26; i++) { if(root->next[i]!=NULL) dealtrie((root->next[i])); } free(root); return 0;}int main(){ int i,m,n,sum; char s1[20000][100],s[100]; while(~scanf("%d",&n),n) { scanf("%d",&m); for(i=0; i<n; i++) { scanf("%s",s1[i]); } struct code *root=newnode(); for(i=0; i<m; i++) { scanf("%s",s); insert(root,s); } sum=0; for(i=0; i<n; i++) { if(find(root,s1[i])) sum++; } printf("%d\n",n-sum); dealtrie(root); } return 0;}
0 0
- 字典树sdutoj 1500模板
- SDUTOJ 1500 ——Message Flood 字典树或map
- SDUTOJ 3039 迷之好奇 静态字典树
- SDUTOJ 1500
- 字典树 模板
- 字典树【模板】
- 字典树模板
- 字典树模板
- 字典树模板
- 经典字典树模板
- hdu1247 字典树模板
- 字典树模板
- 字典树模板
- 字典树模板
- 字典树模板
- 字典树 模板
- 1251 字典树 模板
- 数据结构 字典树模板
- uva-572 - Oil Deposits
- dom4j操作xml的练习
- Unable to handle kernel paging request at virtual address 0x7e005070 解决办法
- 循环的角度求均值
- OpenCV基础篇之使用CMake管理工程
- 字典树sdutoj 1500模板
- XDOJ1180 - 对称数
- OpenCV基础篇之查找表
- 图像二值化
- 找出二叉树中和为某一定值的所有路径
- docker学习 主流的pass平台(体验的是设计模式)
- @Override错误
- VS 自带打包程序
- CentOS上编译安装OpenCV-2.3.1与ffmpeg-2.1.2