PAT(甲)1022 Digital Library (30)（30 分）

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本文介绍了一个基于图书信息的检索系统实现方法，该系统能够根据读者提供的查询条件，如书名、作者、关键词等，返回符合条件的图书ID列表，并按照ID进行升序排序。

1022 Digital Library (30)（30 分）

题目描述：

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.

输入格式：
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

输出格式：
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.

题目大意：
题目就是输入书籍的信息，然后按照要求查找符合搜索条件的相应的书籍ID，并按ID升序输出

解题方法：
这里我是用一个结构体，然后把保存的对象按ID升序排列，最后就是按条件的一个简单查找过程，题目逻辑比较简单，但需要注意几个易错点。

易错点：
1. 输出ID不足7位前面需要补0
2. 对于结构体数组的创建的声明，不能写在主函数中，否则最后一个测试样例会通不过

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++)
    {   /* 输入存储信息 */
        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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