Uva 642 - Word Amalgamation sort qsort
| Word Amalgamation |
In millions of newspapers across the United States there is a word game called Jumble. The object of this game is to solve a riddle, but in order to find the letters that appear in the answer it is necessary to unscramble four words. Your task is to write a program that can unscramble words.
Input
The input file contains four parts:
- 1.
- a dictionary, which consists of at least one and at most 100 words, one per line;
- 2.
- a line containing XXXXXX, which signals the end of the dictionary;
- 3.
- one or more scrambled `words' that you must unscramble, each on a line by itself; and
- 4.
- another line containing XXXXXX, which signals the end of the file.
All words, including both dictionary words and scrambled words, consist only of lowercase English letters and will be at least one and at most six characters long. (Note that the sentinel XXXXXX contains uppercaseX's.) The dictionary is not necessarily in sorted order, but each word in the dictionary is unique.
Output
For each scrambled word in the input, output an alphabetical list of all dictionary words that can be formed by rearranging the letters in the scrambled word. Each word in this list must appear on a line by itself. If the list is empty (because no dictionary words can be formed), output the line ``NOT A VALID WORD" instead. In either case, output a line containing six asterisks to signal the end of the list.
Sample Input
tarp
given
score
refund
only
trap
work
earn
course
pepper
part
XXXXXX
resco
nfudre
aptr
sett
oresuc
XXXXXX
Sample Output
score
******
refund
******
part
tarp
trap
******
NOT A VALID WORD
******
course
******
sort代码
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
char word[100][7];
char* dic[100];
char* seq[100];
char str[7]; bool compare(char *p1, char *p2)
{
return strcmp(p1,p2)<0;
}
int main()
{
int n ;
for (n = 0; scanf("%s", word[n]) && strcmp(word[n], "XXXXXX") != 0; n++);
for (int i=0; i < n; i++)
dic[i] = word[i];
sort(dic,dic+n,compare);
for (int i = 0; i < n; i++)
{
seq[i] = (char *)malloc(strlen(dic[i])+1);
strcpy(seq[i],dic[i]);
sort(seq[i],seq[i]+strlen(seq[i]));
}
while (scanf("%s", str) && strcmp(str, "XXXXXX") != 0)
{
int found = 0;
sort(str,str+strlen(str));
for (int i = 0; i < n;i++)
if (strcmp(str, seq[i]) == 0)
{
found = 1;
printf("%s\n",dic[i]);
}
if (found == 0)
printf("NOT A VALID WORD\n");
printf("******\n");
}
return 0;
}
用qsort代码
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
char word[100][7];
char seq[100][7];
char str[7];
int comp_char(const void * p1, const void * p2)
{
return *(char *)p1 - *(char *)p2;
}
int compare(const void *p1, const void *p2)
{
return strcmp((char *)p1, (char *)p2);
}
int main()
{
int n;
for (n = 0; scanf("%s", word[n]) && strcmp(word[n], "XXXXXX") != 0; n++);
qsort(word,n,sizeof(word[0]),compare);
for (int i = 0; i < n; i++)
{
strcpy(seq[i], word[i]);
qsort(seq[i], strlen(seq[i]),sizeof(char),comp_char);
}
while (scanf("%s", str) && strcmp(str, "XXXXXX") != 0)
{
int found = 0;
qsort(str, strlen(str),sizeof(char),comp_char);
for (int i = 0; i < n; i++)
if (strcmp(str, seq[i]) == 0)
{
found = 1;
printf("%s\n", word[i]);
}
if (found == 0)
printf("NOT A VALID WORD\n");
printf("******\n");
}
return 0;
}
qsort可以对2维数组进行排序,因为它可以调换任意大小的内存块的顺序(第三个参数指定),sort只能对一维数组或者容器进行排序,因为对其解引用时必须是一个左值。
另一个注意点就是cmp函数比较时qsort用“-”,而sort用”>”。
sort函数要求比较函数是strict weak ordering的,否则会出现assertion error: Invalid operator<,而strict weak ordering必须满足三个条件:
1) Strict: pred (X, X) is always false. X跟X自己比为false
2) Weak: If ! pred (X, Y) && !pred (Y, X), X==Y. 当X<Y和Y<X都不成立时,X等于Y
3)Ordering: If pred (X, Y) && pred (Y, Z), then pred (X, Z). 当X<Y,Y<Z时,X<Z成立,即排序的一个传递性。
Uva 642 - Word Amalgamation sort qsort的更多相关文章
- UVa 642 - Word Amalgamation
题目:给你一个单词列表.再给你一些新的单词.输出列表中又一次排列能得到此新单词的词. 分析:字符串.对每一个字符串的字母排序生成新的传f(str).总体排序,用二分来查找就可以. 说明:注意输出要满足 ...
- poj1318 Word Amalgamation 字符串排序(qsort)
Word Amalgamation Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 9794 Accepted: 4701 ...
- Word Amalgamation(枚举 + 排序)
Word Amalgamation Time Limit: 1 Sec Memory Limit: 64 MB Submit: 373 Solved: 247 Description In mil ...
- hdu-----(1113)Word Amalgamation(字符串排序)
Word Amalgamation Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others ...
- Word Amalgamation
Problem Description In millions of newspapers across the United States there is a word game called J ...
- Word Amalgamation(hdoj1113)
Word Amalgamation Problem Description In millions of newspapers across the United States there is a ...
- hdu1113 Word Amalgamation(详解--map和string的运用)
版权声明:本文为博主原创文章.未经博主同意不得转载. vasttian https://blog.csdn.net/u012860063/article/details/35338617 转载请注明出 ...
- HDOJ.1113 Word Amalgamation(map)
Word Amalgamation 点我挑战题目 点我一起学习STL-MAP 题意分析 给出字典.之后给出一系列======乱序======单词,要求你查字典,如过这个乱序单词对用有多个有序单词可以输 ...
- poj 1318 Word Amalgamation
Word Amalgamation Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 9968 Accepted: 4774 ...
随机推荐
- Java字符串==和equals的区别
首先我们来了解一下String类,Java的字符串是一旦被赋值之后无法更改的(这里的无法更改是指不能将字符串中单个或一段字符重新赋值),这也是Java虚拟机为了减少内存开销,避免字符串的重复创建设立的 ...
- 【LiteOS】LiteOS消息队列
目录 前言 链接 参考 笔录草稿 基本概念 队列运作机制 队列运作原理 消息队列传输方式 消息队列的阻塞机制 出队阻塞 入队阻塞 任务相关函数 任务开发流程 注意事项 * 实战 前言 链接 LiteO ...
- (十四--十五)数据库查询优化Part I
(十四--十五)数据库查询优化Part I 如果理解的有问题.欢迎大家指出.这也是我在看课记得笔记..可能会有很多问题 查询优化的重要性 请记住用户只会告诉DMBS他们想要什么样的结果,而不会告诉他们 ...
- Android 之 手动创建活动
•活动是什么 活动(Activity)是最容易吸引用户的地方,它是一种可以包含用户界面的组件: 主要用于和用户进行交互: 一个应用程序可以包含零个或多个活动. 接下来,我们来学习一下活动的基本用法. ...
- 字节、位、bit、byte、KB、B、字符之间的关系
一.bit 位bit就是位,也叫比特位,是计算机表示数据最小的单位,例如 1b ,2b,3b..... 说白了就是0或者1:计算机内存中的存储都是01这两个东西. 二.byte 字节byte就是字节 ...
- Java(25-40)【数据类型转换、运算符、方法入门】
1.ASCII编码表 0--48 A--65 a--97 2. Unicode万国码 字符'中'为20013 3.算数运算符 double类型的加上int类型结果为double byte short ...
- JMeter线程组编辑区揭秘
线程组编辑区如下: 有点复杂,但是慢慢看下来,还是比较容易理解. Name 带有业务含义的名字. Comments 线程组的备注说明. Action to be taken after a Sampl ...
- Unity 协程(Coroutine)原理与用法详解
前言: 协程在Unity中是一个很重要的概念,我们知道,在使用Unity进行游戏开发时,一般(注意是一般)不考虑多线程,那么如何处理一些在主任务之外的需求呢,Unity给我们提供了协程这种方式 为啥在 ...
- golang channel的行为
1. 读nil的channel是永远阻塞的.关闭nil的channel会造成panic. 2. closed channel的行为: (1)向close的channel发消息会panic,关闭一个已经 ...
- 转载:Windows使用tail -f 监控文件
https://www.cnblogs.com/my-bambi/p/11793770.html