POJ 2001-Shortest Prefixes【字典树】
2016-05-16 18:18
423 查看
Shortest Prefixes
Description
A prefix of a string is a substring starting at the beginning of the given string. The prefixes of "carbon" are: "c", "ca", "car", "carb", "carbo", and "carbon". Note that the empty string is not considered a prefix in this problem, but every non-empty string
is considered to be a prefix of itself. In everyday language, we tend to abbreviate words by prefixes. For example, "carbohydrate" is commonly abbreviated by "carb". In this problem, given a set of words, you will find for each word the shortest prefix that
uniquely identifies the word it represents.
In the sample input below, "carbohydrate" can be abbreviated to "carboh", but it cannot be abbreviated to "carbo" (or anything shorter) because there are other words in the list that begin with "carbo".
An exact match will override a prefix match. For example, the prefix "car" matches the given word "car" exactly. Therefore, it is understood without ambiguity that "car" is an abbreviation for "car" , not for "carriage" or any of the other words in the list
that begins with "car".
Input
The input contains at least two, but no more than 1000 lines. Each line contains one word consisting of 1 to 20 lower case letters.
Output
The output contains the same number of lines as the input. Each line of the output contains the word from the corresponding line of the input, followed by one blank space, and the shortest prefix that uniquely (without ambiguity) identifies this word.
Sample Input
Sample Output
Source
Rocky Mountain 2004
解题思路:
输入完上面的所有字符后我们要 ctrl+z,输出每个字符的最长序列,只要这里面又一次出现就算。
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 16880 | Accepted: 7327 |
A prefix of a string is a substring starting at the beginning of the given string. The prefixes of "carbon" are: "c", "ca", "car", "carb", "carbo", and "carbon". Note that the empty string is not considered a prefix in this problem, but every non-empty string
is considered to be a prefix of itself. In everyday language, we tend to abbreviate words by prefixes. For example, "carbohydrate" is commonly abbreviated by "carb". In this problem, given a set of words, you will find for each word the shortest prefix that
uniquely identifies the word it represents.
In the sample input below, "carbohydrate" can be abbreviated to "carboh", but it cannot be abbreviated to "carbo" (or anything shorter) because there are other words in the list that begin with "carbo".
An exact match will override a prefix match. For example, the prefix "car" matches the given word "car" exactly. Therefore, it is understood without ambiguity that "car" is an abbreviation for "car" , not for "carriage" or any of the other words in the list
that begins with "car".
Input
The input contains at least two, but no more than 1000 lines. Each line contains one word consisting of 1 to 20 lower case letters.
Output
The output contains the same number of lines as the input. Each line of the output contains the word from the corresponding line of the input, followed by one blank space, and the shortest prefix that uniquely (without ambiguity) identifies this word.
Sample Input
carbohydrate cart carburetor caramel caribou carbonic cartilage carbon carriage carton car carbonate
Sample Output
carbohydrate carboh cart cart carburetor carbu caramel cara caribou cari carbonic carboni cartilage carti carbon carbon carriage carr carton carto car car carbonate carbona
Source
Rocky Mountain 2004
解题思路:
输入完上面的所有字符后我们要 ctrl+z,输出每个字符的最长序列,只要这里面又一次出现就算。
#include<stdio.h> #include<string.h> #include<algorithm> #include<queue> using namespace std; typedef struct node { int cnt; node *next[26]; }Trie; Trie root; void ct(char *str) { int len,i,j; len=strlen(str); Trie *p=&root,*q; for(i=0;i<len;i++) { int id=str[i]-'a'; if(p->next[id]==NULL) { q=new Trie; q->cnt=1; for(j=0;j<26;j++) q->next[j]=NULL; p->next[id]=q; p=p->next[id]; } else { p->next[id]->cnt++; p=p->next[id]; } } } void Find_Trie(char*str) { int len,i,j,id; len=strlen(str); Trie*p=&root; for( i=0; i<len; i++){ id=str[i]-'a'; if( p->next[id]==NULL) return ; if(p->next[id]->cnt==1) { printf("%c",str[i]); return ; } else if(p->next[id]->cnt>1) { printf("%c",str[i]); p=p->next[id]; } } return ; } char xx[1050][25]; int main() { char map[1050]; int ii=0; while(scanf("%s",map)!=EOF) { ct(map); strcpy(xx[ii],map); ii++; } for(int j=0;j<ii;j++) { printf("%s ",xx[j]); Find_Trie(xx[j]); printf("\n"); } return 0; }
相关文章推荐
- Android Studio运行程序时显示apk not exists
- 嵌入式linux的网络编程(2)--TCP Server程序设计
- div的显示
- Linux常用命令汇总-mv
- synchronized(this)和synchronized(object)区别
- 安装SQL Server 2008 哪些功能需要选择?
- [设计模式]之六大设计原则
- uWSGI uwsgi_response_write_body_do(): Connection reset by peer 报错的解决方法
- Vim简明教程
- 在VMware ESXi中使用固态硬盘
- 在VMware ESXi中使用固态硬盘
- latex 基本用法(五)
- Linux检查图形化界面是否安装
- 关于个别网页打不开,的问题分析及解决方法
- AlterDialog简单使用一
- 使用XHProf查找PHP性能瓶颈
- HUD 2089 位数dp
- Node异步执行&回调函数
- redis单机的搭建
- C语言中关于除法和取余的理解