Immediate Decodability(字典树-判断是否存在前缀)
2015-08-30 09:01
656 查看
Immediate Decodability
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2328 Accepted Submission(s): 1205
Problem Description
An encoding of a set of symbols is said to be immediately decodable if no code for one symbol is the prefix of a code for another symbol. We will assume for this problem that all codes are in binary, that no two codes within a set of codes are the same, that
each code has at least one bit and no more than ten bits, and that each set has at least two codes and no more than eight.
Examples: Assume an alphabet that has symbols {A, B, C, D}
The following code is immediately decodable:
A:01 B:10 C:0010 D:0000
but this one is not:
A:01 B:10 C:010 D:0000 (Note that A is a prefix of C)
Input
Write a program that accepts as input a series of groups of records from input. Each record in a group contains a collection of zeroes and ones representing a binary code for a different symbol. Each group is followed by a single separator record containing
a single 9; the separator records are not part of the group. Each group is independent of other groups; the codes in one group are not related to codes in any other group (that is, each group is to be processed independently).
Output
For each group, your program should determine whether the codes in that group are immediately decodable, and should print a single output line giving the group number and stating whether the group is, or is not, immediately decodable.
Sample Input
01 10 0010 0000 9 01 10 010 0000 9
Sample Output
Set 1 is immediately decodable Set 2 is not immediately decodable
Source
Pacific Northwest 1998
#include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> using namespace std; char word[15][20]; struct Tree { int num;//判断此节点是否为空的(就是一棵树的结束点) Tree *next[2];//这个是这棵树的子节点 Tree() { num=0; for(int i=0;i<2;i++) { next[i]=NULL; } }//这是一棵树 }*root;//建立一个树根 void insert(Tree *p,char *s) { int i=0; while(s[i])//当这个字符串的某一个元素不为空的时候 { int x=s[i]-'0';//看这个元素在哪里? if(p->next[x]==NULL)//然后看这棵树有没有这个元素如果没有 { p->next[x]=new Tree();//就在这棵树的此节点重新建立一棵子树 } p=p->next[x];//p指向他的子树,这一点你就可以知道了所有字符串的元素原来他们都是主仆关系没有相同的等级关系! p->num++; i++;//继续遍历 } }//插入字符 int find(Tree *p,char *s) { int i=0,ans=0;//寻找树的元素,ans是结果! while(s[i])//当他不为空的时候就是字符串还有的时候 { int n=s[i]-'0';//找到他在这个树的节点 if(p->next )//如果节点不为空 { p=p->next ;//p就指向该节点 ans=p->num;//结果应该为1 i++; } else { return 0; } } return ans; } int deltree(Tree* T) { int i; if(T==NULL) return 0; for(i=0;i<2;i++) { if(T->next[i]!=NULL) deltree(T->next[i]); } free(T); return 0; } int main() { int times=1,i=0,j,success=1; char s[20]; root=new Tree(); while(cin>>s) { if(strcmp(s,"9")==0) { for(j=0;j<i;j++) { // cout<<find(root,word[j])<<endl; if(find(root,word[j])>1) { success=0; break; } } if(success) { printf("Set %d is immediately decodable\n",times++); } else { printf("Set %d is not immediately decodable\n",times++); } deltree(root); root=new Tree(); success=1; i=0; } else { strcpy(word[i],s); insert(root,word[i]); i++; } } }
相关文章推荐
- 【J2SE】Java简介
- java多线程
- 值保留原则
- 版本新特性(在程序启动的时候判断是否是新版本)
- SQL---触发器
- 作业:两个数的正差值
- Access建表SQL语句Create Table设置自动增长列的关键字AUTOINCREMENT使用方法
- Android中的DatePickerDiaolog的使用
- django 1.8 官方文档翻译: 3-4-5 内建基于类的视图的API
- Postgres数据库忘记密码,三个步骤解决(windows下)
- CF-CHAPD-山理山科友谊赛-I
- iOS开发 - 手势移除控制器
- Chef and Prime Divisors Problem code: CHAPD (GCD问题)
- 15 3Sum
- IE6/7/8如何兼容CSS3属性
- 零基础学python-9.4 对象的真值
- 零基础学python-9.4 对象的真值
- win10怎么关闭ie11浏览器?win10关闭ie11浏览器教程
- 电脑双屏显示器小工具UltraMon
- JVM学习—内存方区域与内存溢出异常