uva 123 Searching Quickly
| Searching Quickly |
Background
Searching and sorting are part of the theory and practice of computer science. For example, binary search provides a good example of an easy-to-understand algorithm with sub-linear complexity. Quicksort is an efficient
[average case] comparison based sort.
KWIC-indexing is an indexing method that permits efficient ``human search'' of, for example, a list of titles.
The Problem
Given a list of titles and a list of ``words to ignore'', you are to write a program that generates a KWIC (Key Word In Context) index of the titles. In a KWIC-index, a title is listed once for each keyword that occurs in the title. The KWIC-index is alphabetized by keyword.
Any word that is not one of the ``words to ignore'' is a potential keyword.
For example, if words to ignore are ``the, of, and, as, a'' and the list of titles is:
Descent of Man
The Ascent of Man
The Old Man and The Sea
A Portrait of The Artist As a Young Man
A KWIC-index of these titles might be given by:
a portrait of the ARTIST as a young man
the ASCENT of man
DESCENT of man
descent of MAN
the ascent of MAN
the old MAN and the sea
a portrait of the artist as a young MAN
the OLD man and the sea
a PORTRAIT of the artist as a young man
the old man and the SEA
a portrait of the artist as a YOUNG man
The Input
The input is a sequence of lines, the string :: is used to separate the list of words to ignore from the list of titles. Each of the words to ignore appears in lower-case letters on a line by itself and is no more than 10 characters in length. Each title appears on a line by itself and may consist of mixed-case (upper and lower) letters. Words in a title are separated by whitespace. No title contains more than 15 words.
There will be no more than 50 words to ignore, no more than than 200 titles, and no more than 10,000 characters in the titles and words to ignore combined. No characters other than 'a'-'z', 'A'-'Z', and white space will appear in the input.
The Output
The output should be a KWIC-index of the titles, with each title appearing once for each keyword in the title, and with the KWIC-index alphabetized by keyword. If a word appears more than once in a title, each instance is a potential keyword.
The keyword should appear in all upper-case letters. All other words in a title should be in lower-case letters. Titles in the KWIC-index with the same keyword should appear in the same order as they appeared in the input file. In the case where multiple instances of a word are keywords in the same title, the keywords should be capitalized in left-to-right order.
Case (upper or lower) is irrelevant when determining if a word is to be ignored.
The titles in the KWIC-index need NOT be justified or aligned by keyword, all titles may be listed left-justified.
Sample Input
is
the
of
and
as
a
but
::
Descent of Man
The Ascent of Man
The Old Man and The Sea
A Portrait of The Artist As a Young Man
A Man is a Man but Bubblesort IS A DOG
Sample Output
a portrait of the ARTIST as a young man
the ASCENT of man
a man is a man but BUBBLESORT is a dog
DESCENT of man
a man is a man but bubblesort is a DOG
descent of MAN
the ascent of MAN
the old MAN and the sea
a portrait of the artist as a young MAN
a MAN is a man but bubblesort is a dog
a man is a MAN but bubblesort is a dog
the OLD man and the sea
a PORTRAIT of the artist as a young man
the old man and the SEA
a portrait of the artist as a YOUNG man
题目大意:给出一些ingore word(可忽略单词),再给出一些句子,句子由可忽略单词和不可忽略单词组成, 要求找出所有的不可忽略单词,输出含不可忽略单词的句子(此时不可忽略单词要大写),注意一个不可忽略单词可能出现在多个句子里,一个句子可能有多个(包括相同的)不可忽略单词(输出时按照不可忽略单词的字典序,相同单词按照句子的顺序)
解题思路:读入不可忽略单词,再读入句子,每读入一个句子时将句子分解成单词,分解的同时判断它是否为不可忽略单词(与前面可忽略单词进行比较),读完所有的句子以后对不可忽略单词进行排序,然后再对每个不可忽略单词查找每条语句。
PS:题目本身没有难度,但是要细心,因为比较繁琐。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std; #define N 10005
#define T 205
#define M 20 struct say{
char word[M];
}s[T]; struct talk{
char sen[T][M];
int cnt;
}title[T]; int n_s = 0, n_title = 0, n_ignore = 0;;
char ignore_word[T][M]; void change(char str[]){
int len = strlen(str);
for (int i = 0; i < len; i++){
if(str[i] >= 'A' && str[i] <= 'Z')
str[i] += 32;
}
} int cmp(const say a, const say b){
return strcmp(a.word, b.word) < 0;
} void print(talk t, int k){
for (int i = 0; i < t.cnt - 1; i++){
if (i == k){
for (int j = 0; j < strlen(t.sen[i]); j++)
printf("%c", t.sen[i][j] - 32);
printf(" ");
}
else
printf("%s ", t.sen[i]);
}
if (t.cnt - 1 == k){
for (int j = 0; j < strlen(t.sen[t.cnt - 1]); j++)
printf("%c", t.sen[t.cnt - 1][j] -32);
printf("\n");
}
else
printf("%s\n", t.sen[t.cnt - 1]);
} void find(talk t, say a){
for (int i = 0; i < t.cnt; i++){
if (strcmp(t.sen[i], a.word) == 0)
print(t, i);
}
} void judge(char str[]){
for (int i = 0; i < n_ignore; i++){
if (strcmp(ignore_word[i], str) == 0)
return;
}
for (int i = 0; i < n_s; i++){
if (strcmp(s[i].word, str) == 0)
return;
}
strcpy(s[n_s++].word, str);
} void build(char str[], int k){
int len = strlen(str), n = 0, m = 0;
for (int i = 0; i < len; i++){
if (str[i] >= 'a' && str[i] <= 'z')
title[k].sen[n][m++] = str[i];
else{
title[k].sen[n][m] = '\0';
judge(title[k].sen[n]);
m = 0;
n++;
}
}
title[k].sen[n++][m] = '\0';
judge(title[k].sen[n - 1]);
title[k].cnt = n;
} int main(){
char name[N]; // Read.
while (1){
gets(ignore_word[n_ignore]);
if (strcmp(ignore_word[n_ignore], "::") == 0)
break;
change(ignore_word[n_ignore]);
n_ignore++;
}
while (gets(name) != NULL){
change(name);
build(name, n_title);
n_title++;
} // Ready.
sort(s, s + n_s, cmp); // Find.
for (int i = 0; i < n_s; i++)
for (int j = 0; j < n_title; j++)
find(title[j], s[i]);
return 0;}
uva 123 Searching Quickly的更多相关文章
- STL --- UVA 123 Searching Quickly
UVA - 123 Searching Quickly Problem's Link: http://acm.hust.edu.cn/vjudge/problem/viewProblem.acti ...
- UVa 12505 Searching in sqrt(n)
传送门 一开始在vjudge上看到这题时,标的来源是CSU 1120,第八届湖南省赛D题“平方根大搜索”.今天交题时CSU突然跪了,后来查了一下看哪家OJ还挂了这道题,竟然发现这题是出自UVA的,而且 ...
- uva 1597 Searching the Web
The word "search engine" may not be strange to you. Generally speaking, a search engine se ...
- 湖南省第八届大学生程序设计大赛原题 D - 平方根大搜索 UVA 12505 - Searching in sqrt(n)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=30746#problem/D D - 平方根大搜索 UVA12505 - Searchin ...
- UVa123 - Searching Quickly
题目地址:点击打开链接 C++代码: #include <iostream> #include <set> #include <map> #include < ...
- Volume 1. Sorting/Searching(uva)
340 - Master-Mind Hints /*读了老半天才把题读懂,读懂了题输出格式没注意,结果re了两次. 题意:先给一串数字S,然后每次给出对应相同数目的的一串数字Si,然后优先统计Si和S ...
- 刘汝佳 算法竞赛-入门经典 第二部分 算法篇 第五章 3(Sorting/Searching)
第一题:340 - Master-Mind Hints UVA:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Item ...
- UVA题目分类
题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics ...
- UVA题解二
UVA题解二 UVA 110 题目描述:输出一个Pascal程序,该程序能读入不多于\(8\)个数,并输出从小到大排好序后的数.注意:该程序只能用读入语句,输出语句,if语句. solution 模仿 ...
随机推荐
- shell中的数学运算
shell中要进行数学运算通常有3中方法: expr命令 比如 expr 1 + 6就会返回7,使用expr的缺点就是碰到乘法运算,或者加括号(因为它们在shell中有其他意义),需要使用转义,比如: ...
- python-整理-logging日志
python的日志功能模块是logging 功能和使用方式非常类似于log4 如何使用logging: # 导入日志模块import logging# 使用配置文件设置日志时,需要导入这个模块 imp ...
- Array类型(一)
1.创建数组的基本方式有两种,第一种使用Array构造函数 使用Array构造函数时可以省略new操作符 2.第二种方式是使用数组字面量表示法 由于IE的实现与其他浏览器不一致,因此我们不赞同使用这个 ...
- 异常处理与调试5 - 零基础入门学习Delphi54
调试(Debug) 让编程改变世界 Change the world by program [caption id="attachment_2731" align="al ...
- Wireless Network(POJ 2236)
Wireless Network Time Limit: 10000MS Memory Limit: 65536K Total Submissions: 20724 Accepted: 871 ...
- MVC中配置OutputCache的VaryByParam参数无效的问题
在项目使用OutputCacheAttribute是遇到了问题,当我想在配置文件web.config中配置OutputCache的VaryByParam时竟然不起作用,下面是相关代码: 文件FaceC ...
- jquery.fn.extend与jquery.extend(转)
jQuery为开发插件提拱了两个方法,分别是: JavaScript代码 jQuery.fn.extend(object); jQuery.extend(object); jQuery.extend( ...
- Android IntentService 与Alarm开启任务关闭任务
1:MyService public class MyService extends IntentService{ AlarmManager alarmManager = null; PendingI ...
- SD card技术了解并WINCE下SDHC驱动开发(updated)
Suumary: 简单介绍了一下SD卡的历史和发展,同时结合MX31 ADS上的WINCE 下SDHC驱动更深入的了解该硬件的一些行为特点. 了解SD card SD是Secure Digital C ...
- 将窗体显示在 PageControl 上
var AWinControl:TPageControl; begin AWinControl := PageControl1; if frmAbout = nil then Exit; frmAbo ...