Description
Background from Wikipedia: "Set theory is a branch of mathematics created principally by the German mathematician Georg Cantor at the end of the 19th century. Initially controversial, set theory has come to play the role of a foundational theory in modern mathematics, in the sense of a theory invoked to justify assumptions made inmathematics concerning the existence of mathematical objects (such as numbers or functions) and their properties. Formal versions of set theory also have a foundational role to play as specifying a theoretical ideal of mathematical rigor in proofs."
Given this importance of sets, being the basis of mathematics, a set
of eccentric theorist set off to construct a supercomputer operating on
sets instead of numbers. The initial Set-Stack Alpha is under
construction, and they need you to simulate it in order to verify the
operation of the prototype.
The computer operates on a single stack of sets, which is initially
empty. After each operation, the cardinality of the topmost set on the
stack is output. The cardinality of a set S is denoted |S| and is the
number of elements in S. The instruction set of the SetStack Alpha is
PUSH, DUP, UNION, INTERSECT, and ADD.
- PUSH will push the empty set {} on the stack.
- DUP will duplicate the topmost set (pop the stack, and then push that set on the stack twice).
- UNION will pop the stack twice and then push the union of the two sets on the stack.
- INTERSECT will pop the stack twice and then push the intersection of the two sets on the stack.
- ADD will pop the stack twice, add the first set to the second one, and then push the resulting set on the stack.
For illustration purposes, assume that the topmost element of the stack is
A = {{}, {{}}}
and that the next one is
B = {{}, {{{}}}}.
For these sets, we have |A| = 2 and |B| = 2. Then:
- UNION would result in the set { {}, {{}}, {{{}}} }. The output is 3.
- INTERSECT would result in the set { {} }. The output is 1.
- ADD would result in the set { {}, {{{}}}, {{},{{}}} }. The output is 3.
Input
An
integer 0 ≤ T ≤ 5 on the first line gives the cardinality of the set of
test cases. The first line of each test case contains the number of
operations 0 ≤ N ≤ 2 000. Then follow N lines each containing one of the
five commands. It is guaranteed that the SetStack computer can execute
all the commands in the sequence without ever popping an empty stack.
Output
For
each operation specified in the input, there will be one line of output
consisting of a single integer. This integer is the cardinality of the
topmost element of the stack after the corresponding command has
executed. After each test case there will be a line with *** (three
asterisks).
Sample Input
2 9 PUSH DUP ADD PUSH ADD DUP ADD DUP UNION 5 PUSH PUSH ADD PUSH INTERSECT
Sample Output
0 0 1 0 1 1 2 2 2 *** 0 0 1 0 0 ***
解析：运用STL可方便地解决这个问题。用vector来存储各个不同的集合，运用map将集合映射为vector中相应的下标。集合的操作UNION、INTERSECT、ADD可分别用algorithm中的set_union、set_intersection以及set自带的insert完成，按题意运用stack模拟栈操作即可。
1 #include <iostream>
2 #include <set>
3 #include <map>
4 #include <vector>
5 #include <stack>
6 #include <algorithm>
7 using namespace std;
8
9 map<set<int>, int> id;
10 map<set<int>, int>::iterator it;
11 vector<set<int> > v;
12
13 int getID(set<int> x)
14 {
15 it = id.find(x);
16 if(it != id.end())
17 return it->second;
18 v.push_back(x);
19 return id[x] = v.size()-1;
20 }
21
22 int main()
23 {
24 int T, n;
25 cin>>T;
26 while(T--){
27 cin>>n;
28 id.clear();
29 v.clear();
30 stack<int> s;
31 string op;
32 while(n--){
33 cin>>op;
34 if(op[0] == ‘P‘)
35 s.push(getID(set<int>()));
36 else if(op[0] == ‘D‘)
37 s.push(s.top());
38 else{
39 set<int> t1 = v[s.top()]; s.pop();
40 set<int> t2 = v[s.top()]; s.pop();
41 set<int> t;
42 if(op[0] == ‘U‘){
43 set_union(t1.begin(), t1.end(), t2.begin(), t2.end(), inserter(t, t.begin()));
44 s.push(getID(t));
45 }
46 else if(op[0] == ‘I‘){
47 set_intersection(t1.begin(), t1.end(), t2.begin(), t2.end(), inserter(t, t.begin()));
48 s.push(getID(t));
49 }
50 else if(op[0] == ‘A‘){
51 t2.insert(getID(t1));
52 s.push(getID(t2));
53 }
54 }
55 cout<<v[s.top()].size()<<endl;
56 }
57 cout<<"***"<<endl;
58 }
59 return 0;
60 }