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("
");
	}
	else
		printf("%s
", 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] = '';
			judge(title[k].sen[n]);
			m = 0;
			n++;
		}
	}
	title[k].sen[n++][m] = '';
	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;}