1022 Digital Library (30)（30 分）
題目描述：
A Digital Library contains millions of books, stored according to their titles, authors, key words of their abstracts, publishers, and published years. Each book is assigned an unique 7-digit number as its ID. Given any query from a reader, you are supposed to output the resulting books, sorted in increasing order of their ID’s.
- 輸入格式：
Each input file contains one test case. For each case, the first line contains a positive integer N (<=10000) which is the total number of books. Then N blocks follow, each contains the information of a book in 6 lines: - Line #1: the 7-digit ID number;
- Line #2: the book title – a string of no more than 80 characters;
- Line #3: the author – a string of no more than 80 characters;
- Line #4: the key words – each word is a string of no more than 10
characters without any white space, and the keywords are separated by
exactly one space; - Line #5: the publisher – a string of no more than 80 characters;
- Line #6: the published year – a 4-digit number which is in the range
[1000, 3000].
It is assumed that each book belongs to one author only, and contains no more than 5 key words; there are no more than 1000 distinct key words in total; and there are no more than 1000 distinct publishers.
After the book information, there is a line containing a positive integer M (<=1000) which is the number of user’s search queries. Then M lines follow, each in one of the formats shown below:
1: a book title
2: name of an author
3: a key word
4: name of a publisher
5: a 4-digit number representing the year
- 輸出格式：
For each query, first print the original query in a line, then output the resulting book ID’s in increasing order, each occupying a line. If no book is found, print “Not Found” instead.
題目大意：
題目就是輸入書籍的信息，然後按照要求查找符合搜索條件的相應的書籍ID，並按ID升序輸出
解題方法：
這裏我是用一個結構體，然後把保存的對象按ID升序排列，最後就是按條件的一個簡單查找過程，題目邏輯比較簡單，但需要注意幾個易錯點。
易錯點：
1. 輸出ID不足7位前面需要補0
2. 對於結構體數組的創建的聲明，不能寫在主函數中，否則最後一個測試樣例會通不過
程序：
#include <cstdio>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
struct Info
{
char title[200], autor[200], key[200], pub[200], year[200];
int ID;
}infos[10001];
bool cmp(Info in1, Info in2)
{
return in1.ID < in2.ID;
}
int main(int argc, char const *argv[])
{
int N, M, search, flag;
char searchInfo[200];
scanf("%d", &N);
for (int i = 0; i < N; i++)
{ /* 輸入存儲信息 */
scanf("%d", &infos[i].ID);
getchar();
fgets(infos[i].title, 200, stdin); infos[i].title[strlen(infos[i].title) - 1] = '\0';
fgets(infos[i].autor, 200, stdin); infos[i].autor[strlen(infos[i].autor) - 1] = '\0';
fgets(infos[i].key, 200, stdin); infos[i].key[strlen(infos[i].key) - 1] = '\0';
fgets(infos[i].pub, 200, stdin); infos[i].pub[strlen(infos[i].pub) - 1] = '\0';
scanf("%s", infos[i].year);
}
sort(infos, infos+N, cmp); /* 按ID排序 */
scanf("%d", &M);
getchar();
for (int k = 0; k < M; k++)
{
scanf("%d: ", &search);
fgets(searchInfo, 200, stdin); searchInfo[strlen(searchInfo) - 1] = '\0';
printf("%d: %s\n", search, searchInfo);
flag = 0;
switch(search)
{
case 1:
for (int i = 0; i < N; i++)
if (strcmp(infos[i].title, searchInfo) == 0){
printf("%07d\n", infos[i].ID);
flag = 1;
}
break;
case 2:
for (int i = 0; i < N; i++)
if (strcmp(infos[i].autor, searchInfo) == 0){
printf("%07d\n", infos[i].ID);
flag = 1;
}
break;
case 3:
for (int i = 0; i < N; i++){
char *c;
c = strstr(infos[i].key, searchInfo);
if (c){
printf("%07d\n", infos[i].ID);
flag = 1;
}
}
break;
case 4:
for (int i = 0; i < N; i++)
if (strcmp(infos[i].pub, searchInfo) == 0){
printf("%07d\n", infos[i].ID);
flag = 1;
}
break;
case 5:
for (int i = 0; i < N; i++)
if (strcmp(infos[i].year, searchInfo) == 0){
printf("%07d\n", infos[i].ID);
flag = 1;
}
break;
}
if (flag == 0)
printf("Not Found\n");
}
return 0;
}
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