题意：
输入若干书籍和作者名字，然后先按作者名字升序排列，再按标题升序排列，然后会有3种指令，BORROW,RETURN, SHELVE。
BORROW 和 RETURN 都会带有一个书名在后面，如：
BORROW "The Canterbury Tales"
RETURN "The Canterbury Tales"
当遇到SHELVE指令，输出已还但没上架的书排序后（规则同上）依次插入书架的序列。
用例：
输入：
"The Canterbury Tales" by Chaucer, G.
"Algorithms" by Sedgewick, R.
"The C Programming Language" by Kernighan, B. and Ritchie, D.
END
BORROW "The Canterbury Tales"
RETURN "The Canterbury Tales"
BORROW "The C Programming Language"
RETURN "The C Programming Language"
SHELVE
BORROW "Algorithms"
RETURN "Algorithms"
SHELVE
END
输出：
Put "The Canterbury Tales" first
Put "The C Programming Language" after "The Canterbury Tales"
END
Put "Algorithms" after "The C Programming Language"
END
分析：
这题的读入有点麻烦，我的处理是使用getline， 然后将 “ 号变成空格 将空格变成“#”（ASCII差距不大）， 再用stringsteam输入到string中，最后输出再将“#”转换为“ ”。
网上的方法是使用string.substr()截取,用string.find()去寻找位置， 我截取这种方法更好（stringsteam很慢）。
用map<string,int>将string映射到书架位置上面，用一个bool去标记这个位置有没被拿走，注意return后不能直接把书放回去，要标记，等收到SHELVE指令再排序放书。
1 #include <cstdio>
2 #include <cstring>
3 #include <vector>
4 #include <map>
5 #include <string>
6 #include <iostream>
7 #include <set>
8 #include <sstream>
9 #include <algorithm>
10 using namespace std;
11 struct B
12 {
13 string name;
14 string author;
15 B(){}
16 B(const string& _name,const string& _author):
17 name(_name),author(_author){}
18 friend bool operator < (const B a,const B b)
19 {
20 if(a.author < b.author) return true;
21 else
22 {
23 if(a.author == b.author)
24 {
25 if(a.name < b.name)
26 return true;
27 else return false;
28 }
29 else return false;
30 }
31 }
32 };
33 struct A
34 {
35 int a;
36 int b;
37 A(){}
38 A(int _a, int _b):
39 a(_a), b(_b){}
40 // a after b of a is the first(b=-1);
41 };
42 typedef set<B> :: iterator sIT;
43 typedef vector<B> :: iterator vIT;
44 typedef vector<A> :: iterator aIT;
45 typedef vector<int> :: iterator IT;
46 string trans(const string& a)
47 {
48 string t = a;
49 for(int i = 0; i < t.size(); i++)
50 {
51 if(t[i] == ‘#‘) t[i] = ‘ ‘;
52 }
53 return t;
54 }
55 int main()
56 {
57 ios::sync_with_stdio(false);
58 #if LOCAL
59 freopen("1.txt","r",stdin);
60 freopen("2.txt","w",stdout);
61 #endif // LOCAL
62 string s;
63 vector<B> book;
64 bool lend[1005];
65 vector<A> ans;
66 fill(lend, lend+1005,1);
67 map<string,int> id;
68 vector<int> ret;
69 while(getline(cin,s))
70 {
71 for(int i = 0; i < s.size(); i++)
72 {
73 if(s[i] == ‘"‘) s[i] = ‘ ‘;
74 else if(s[i] == ‘ ‘) s[i] = ‘#‘;
75 }
76 stringstream ss(s);
77 string tname, tauthor;
78 if(!(ss>>tname>>tauthor)) break;
79 book.push_back(B(tname,tauthor));
80 }
81 sort(book.begin(), book.end());
82 int cnt = 0;
83 for(vIT it = book.begin(); it != book.end(); it++)
84 {
85 id[(*it).name] = cnt++;
86 }
87 string op;
88 while(cin>>op)
89 {
90 if(op=="END") break;
91 if(op == "SHELVE")
92 {
93 sort(ret.begin(), ret.end());
94 for(IT it = ret.begin(); it != ret.end(); it++)
95 {
96 lend[*it] = 1;
97 int c = (*it) -1;
98 for(; c >= 0 && lend[c] == 0; c--);
99 ans.push_back(A(*it,c));
100 }
101 ret.clear();
102 for(aIT it = ans.begin(); it != ans.end(); it++)
103 {
104 if((*it).b != -1)
105 {
106 cout<<"Put \""<< trans(book[(*it).a].name) << "\" after \"" << trans(book[(*it).b].name) << "\"\n";
107
108 }
109 else
110 {
111 cout<<"Put \""<<trans(book[(*it).a].name)<< "\" first"<<"\n";
112
113 }
114 }
115 ans.clear();
116 cout<<"END\n";
117 continue;
118 }
119 cin.ignore();
120 string b;
121 getline(cin,b);
122 for(int i = 0; i < b.size(); i++)
123 {
124 if(b[i] == ‘"‘) b[i] = ‘ ‘;
125 else if(b[i] == ‘ ‘) b[i] = ‘#‘;
126 }
127 stringstream ss(b);
128 ss>>b;
129 if(op== "BORROW")
130 {
131 lend[id[b]] = 0;
132 }
133 else if(op == "RETURN")
134 {
135
136 ret.push_back(id[b]);
137 }
138 }
139
140
141 return 0;
142 }