UVA - 230 Borrowers string+vector

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题目大意：给出n本书，每本书都有相应的title和author，先按照author的大小排序，然后按title的大小排序
现在给出三种操作
BORROW title：借出书名为tile的书
RETURN title：还书，书名为title
SHELVE：把归还的书放回书架，并输出title和放的位置

解题思路：用一个结构体纪录每本书的状态，然后用一个动态数组存储书，具体看代码，对string这stl不太懂，就参考了别人的代码了。。。

#include<cstdio>#include<cstring>#include<string>#include<iostream>#include<algorithm>#include<vector>using namespace std;struct Book{    string author, title;    bool borrowed, returned;    Book(string a, string b) {        title = a;        author = b;        borrowed = returned = false;    }};bool cmpa(Book a, Book b) {    return a.author < b.author;}bool cmpb(Book a, Book b) {    return a.title < b.title;}vector<Book>  All;string cmd, in, req;void shelve() {    int j;    for(int i = 0; i < All.size(); i++)         if(All[i].returned == true) {            for(j = i; j >= 0; j--)                if(All[j].borrowed == false)                    break;            if(j == -1)                printf("Put %s first\n", All[i].title.c_str());            else                printf("Put %s after %s\n",All[i].title.c_str(), All[j].title.c_str());            All[i].borrowed = All[i].returned = false;        }    cout << "END\n";}void borrow() {    getline(cin,req);    for(int i = 0; i < All.size(); i++)        if(All[i].title == req) {            All[i].borrowed = true;            return ;        }}void back() {    getline(cin,req);    for(int i = 0; i < All.size(); i++)        if(All[i].title == req) {            All[i].returned = true;            return ;        }}int main() {    while(getline(cin,in) && in != "END")         All.push_back(Book(in.substr(0,in.find_last_of("\"") + 1),in.substr(in.find_last_of("\"") + 1) ) );    stable_sort(All.begin(),All.end(),cmpb);    stable_sort(All.begin(),All.end(),cmpa);    while(cin >> cmd)         if(cmd == "BORROW")            cin.get(), borrow();        else if(cmd == "RETURN")            cin.get(), back();        else if(cmd == "SHELVE")            cin.get(), shelve();    return 0;}

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