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 3 Inkfish Henry Carp Max Jericho Carp Max Carp 0
示例输出
3
法1：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char str[20001][11],zancun[11];
struct node{
int flag;
struct node *next[26];//通过结构体数组来指向26个方向
};
struct node *newnode(){
int i;
struct node *p=(struct node *)malloc(sizeof(struct node));此处不要忘了强制转化类型
p->flag=0;标记此时的节点为0
for(i=0;i<26;i++)
p->next[i]=NULL;让26个方向都指空
return p;
}
void insert(struct node *root, char *s)建立字典树，自身调用
{
struct node *p=root;
int i,t,k=strlen(s);
for(i=0;i<k;i++)
{
if(s[i]>='a'&&s[i]<='z')消除输入的字符是大写还是小写的影响
t=s[i]-'a';
else if(s[i]>='A'&&s[i]<='Z')
t=s[i]-'A';
if(p->next[t]==NULL)
p->next[t]=newnode();以此建立节点
p=p->next[t];
}
p->flag=1;名字的最后一个字母进行标记
}
int search(struct node *root, char *s)//查找是否有相同的情况
{
int i,t,k=strlen(s);
struct node *p=root;
for(i=0;i<k;i++)
{
if(s[i]>='a'&&s[i]<='z')
t=s[i]-'a';
else if(s[i]>='A'&&s[i]<='Z')
t=s[i]-'A';
if(p->next[t]==NULL)
return 0;如果节点后面为空，则返回0表示没有
p=p->next[t];
}
if(p->flag==1)如果最后一个标记，则其flag为1， 进入if语句
{
p->flag=0;因为一个人可以给她发多条短信，将标记去掉防止以后重复名字再使总人数减一 return 1;
}
return 0;
}
void Delete(struct node *root)//这是内存释放函数，防止内存过多超内存
{
int i;
for(i = 0 ; i < 26 ; i++)
{
if(root->next[i] != NULL)
Delete(root->next[i]);递归调用自身
}
free(root);
}
int main()
{
int i,n,m,s;
struct node *root;
char b[100];
while(scanf("%d",&n)!=EOF && n!=0)
{
scanf("%d",&m);
s=n;
root=newnode();建立第一个节点
getchar();
for(i=0;i<n;i++)输入n组字符串
{
scanf("%s",str[i]);
}
for(i=0;i<m;i++)
{
scanf("%s",b);
insert (root,b);调用插入函数，建树
}
for(i=0;i<n;i++)
{
if(search(root,str[i]))如果search函数返回值为1，则执行if语句，s--；
s--;
}
printf("%d
",s);
Delete(root);
}
return 0;
}
法2：
#include <stdio.h>
#include <string.h>
struct node
{
int flag;
int z[26];
} num[1000001];
char s[100];
char ss[100];
int ns=0;
int newnode()
{
int p=ns;
ns++;
for(int i=0; i<26; i++)
{
num[p].z[i]=-1;
}
num[p].flag=0;
return p;
}
void creat()
{
int i,k;
scanf("%s",s);
k=strlen(s);
int qian;
qian=0;
int p;
for(i=0; i<k; i++)
{
if(s[i]>='A'&&s[i]<='Z')s[i]=s[i]-'A'+'a';//A-a;
int t=s[i]-'a';
if(num[qian].z[t]==-1)
{
p=newnode();
num[qian].z[t]=p;
}
else p=num[qian].z[t];
qian=p;
}
num[p].flag=1;
}
int su=0;
int len;
int Search(int rt)
{
int j;
int q=rt;
for(j=0; j<len; j++)
{
if(ss[j]>='A'&&ss[j]<='Z')ss[j]=ss[j]-'A'+'a';//A-a;
if(num[q].z[ss[j]-'a']==-1)return 0;
q=num[q].z[ss[j]-'a'];
}
if(num[q].flag==1)
{
num[q].flag=0;
return 1;
}
return 0;
}
int main()
{
int n,m;
while(~scanf("%d",&n))
{
if(n==0)
break;
scanf("%d",&m);
ns=0;
int root=newnode();
int nn=n;
while(n--)
{
creat();
}
int sum=0;
while(m--)
{
scanf("%s",ss);
len=strlen(ss);
sum+=Search(root);
// printf("%d
",sum);
}
printf("%d
",nn-sum);
}
return 0;
}
反思：第一次写的时候Memory Limit Exceeded， 可能很多人遇到这种情况，我开始建树的时候，把电话本中的朋友名单建成了树，其实这样会超内存，可以把给他发短信的同学的名字建成树，也就是说用电话本中的每一个朋友名字，到建的树中去匹配，如果匹配，s--即可，最后输出s。