• 1500_Message Flood


    Message Flood

    Time Limit:1500MS  Memory Limit:65536K
    Total Submit:284 Accepted:40

    Description

    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.

    Input

    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.

    Output

    For each case, print one integer in one line which indicates the number of left friends he must send.

    Sample Input

    5 3
    Inkfish
    Henry
    Carp
    Max
    Jericho
    Carp
    Max
    Carp
    0

    Sample Output


    二分法+快排,试着用哈希表做了,但做了半天没能解决
    View Code
    #include <stdio.h>
    #include
    <string.h>
    #include
    <stdlib.h>
    struct no
    {
    char s[11];
    int f;
    }a[
    20001];
    int cmp(const void * a, const void *b)
    {
    return strcmp((char *)a, (char *)b);
    }
    int main()
    {
    int n,m;
    int i, j, len;
    char s1[11],s2[11];
    while(scanf("%d",&n),n)
    {
    scanf(
    "%d",&m);
    getchar();
    for(i = 0; i < n; i++)
    {
    gets(s1);
    len
    = strlen(s1);
    for(j = 0; j < len; j++)
    if(s1[j] >= 'a')
    s1[j]
    -= 32;
    strcpy(a[i].s,s1);
    a[i].f
    = 1;
    }
    qsort(a,n,
    sizeof(a[0]),cmp);
    for(i = 0; i < m; i++)
    {
    gets(s2);
    len
    = strlen(s2);
    for(j = 0; j < len; j++)
    if(s2[j] >= 'a')
    s2[j]
    -= 32;
    int low = 0, high = n-1, mid;
    while(low <= high)
    {
    mid
    = (low+high)/2;
    if(strcmp(a[mid].s,s2) == 0)
    {
    a[mid].f
    = 0;
    break;
    }
    else if(strcmp(a[mid].s,s2) > 0)
    high
    = mid-1;
    else low = mid+1;
    }
    }
    int num = 0;
    for(i = 0; i < n; i++)
    if(a[i].f)
    num
    ++;
    printf(
    "%d\n",num);
    }
    return 0;
    }

    Trie 树做法:

    View Code
    #include <stdio.h>
    #include
    <string.h>
    #include
    <malloc.h>
    typedef
    struct node
    {
    int flag;
    struct node *next[26];
    }Node;
    int tmp;
    Node
    * NewNode()
    {
    int i;
    Node
    * p = (Node *)malloc(sizeof(Node));
    p
    ->flag = 0;
    for(i = 0; i < 26; i++)
    p
    ->next[i] = 0;
    return p;
    }
    void insert(Node * root, char s[])
    {
    int n = strlen(s);
    int i;
    Node
    * p;
    p
    = root;
    for(i = 0; i < n; i++)
    {
    if(p->next[s[i] - 'a'] == NULL)
    p
    ->next[s[i] - 'a'] = NewNode();
    p
    = p->next[s[i] - 'a'];
    }
    p
    ->flag = 1;
    }
    void search(Node * root, char s[])
    {
    int len = strlen(s);
    int i;
    Node
    *p;
    p
    = root;
    for(i = 0; i < len; i++)
    {
    if(p->next[s[i] - 'a'] != NULL)
    p
    = p->next[s[i] - 'a'];
    }
    if(p->flag)
    {
    p
    ->flag = 0;
    tmp
    --;
    }
    }
    void del(Node * p)
    {
    int i;
    if(p)
    {
    for(i = 0; i < 26; i++)
    if(p->next[i])
    del(p
    ->next[i]);
    }
    free(p);
    p
    = NULL;
    }
    void change(char s[])
    {
    int len, i;
    len
    = strlen(s);
    for(i = 0; i < len; i++)
    if(s[i] >= 'A' && s[i] <= 'Z')
    s[i]
    += 32;
    }
    int main()
    {
    int n, m;
    char s[11];
    Node
    *root;
    while(scanf("%d", &n),n)
    {
    tmp
    = n;
    scanf(
    "%d", &m);
    root
    = NewNode();
    while(n--)
    {
    scanf(
    "%s", s);
    change(s);
    insert(root, s);
    }
    while(m--)
    {
    scanf(
    "%s", s);
    change(s);
    search(root, s);
    }
    printf(
    "%d\n", tmp);
    //del(root);
    }
    return 0;
    }


  • 相关阅读:
    局域网内其他主机如何访问运行在宿主机的虚拟机中的Django项目(转)
    python3脚本获取本机公网ip
    django之允许外部机器访问
    Mysql命令drop database:删除数据库
    CentOS和Windows下配置MySQL远程访问的教程
    Windows 下 MySql 5.7.20安装及data和my.ini文件的配置(转)
    远程登陆linux连接mysql root账号报错:2003-can't connect to MYSQL serve(转)
    Mysql创建用户并授权以及开启远程访问
    Android 之 自定义标签 和 自定义组件
    再探java基础——零碎基础知识整理
  • 原文地址:https://www.cnblogs.com/vongang/p/2114861.html
Copyright © 2020-2023  润新知