UVA 12096 The SetStack Computer（stack及其它STL容器综合使用）

深藏阁楼爱情的钟 2022-06-08 12:28 29阅读 0赞

Background from Wikipedia: “Set theory is a branch of mathematics created principally by the German mathe-matician 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 in mathematics concerning the existence of mathe-matical 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 in-stead of numbers. The initial SetStack 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 eachtest case contains the number of operations 0 ≤ N ≤ 2000. 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  
\*\*\*

**题意：**

对于一个以集合为元素的栈，初始时栈为空。

输入的命令有如下几种：

PUSH：将空集\{\}压栈

DUP：将栈顶元素复制一份压入栈中

UNION：先进行两次弹栈，将获得的集合A和B取并集，将结果压栈

INTERSECTION：先进行两次弹栈，将获得的集合A和B取交集，将结果压栈

ADD：先进行两次弹栈，将获得的集合A和B中，先出栈的集合（如A先）加入到后出栈的集合，将结果压栈

输出每一步操作后栈顶集合的元素的个数。

**思路：**

本题不是简单的整数或字符串集合，而是集合的集合。因此放了方便，可以用STL中的 set<int> 来表示每一个集合，用 map<set<int>,int> 为每一个不同的集合分配一个唯一的ID，就可以用 stack<int> 来表示整个栈。而题目中的各种操作也可以用强大的STL来表示。

**代码：**

#include <iostream>
    #include <algorithm>
    #include <cstdlib>
    #include <map>
    #include <set>
    #include <stack>
    #include <vector>
    #define all(x) x.begin(),x.end()//迭代器
    #define ins(x) inserter(x,x.begin())
    
    using namespace std;
    
    typedef set<int> Set;//用数字ID集合表示题目中的集合
    
    map<Set,int> IDcache;//把集合映射成ID
    vector<Set> setcache;//根据ID取集合
    stack<int> s;
    
    //查找给定集合x的ID，如找不到，分配一个新的ID
    int ID(Set x)
    {
        if(IDcache.count(x))
           return IDcache[x];
        setcache.push_back(x);//添加新集合
        return IDcache[x]=setcache.size()-1;
    }
    
    int main()
    {
        int t;
        while(cin>>t)
        {
            while(t--)
            {
                int n;
                cin>>n;
                while(n--)
                {
                    string cmd;
                    cin>>cmd;
                    if(cmd[0]=='P')
                        s.push(ID(Set()));
                    else if(cmd[0]=='D')
                        s.push(s.top());
                    else
                    {
                        Set x1=setcache[s.top()]; s.pop();
                        Set x2=setcache[s.top()]; s.pop();
                        Set x;
                        if(cmd[0]=='U')
                            set_union(all(x1),all(x2),ins(x));
                        if(cmd[0]=='I')
                            set_intersection(all(x1),all(x2),ins(x));
                        if(cmd[0]=='A')
                        {
                            x=x2;
                            x.insert(ID(x1));
                        }
                        s.push(ID(x));
                    }
                    cout<<setcache[s.top()].size()<<endl;
                }
                cout<<"***"<<endl;
            }
        }
        return 0;
    }