stl(set+stack) LA 3634 The SetStack Computer

题目传送门

题意：给一些对集合的操作，询问每一次操作后栈顶的集合元素个数

分析：首先{}是空的，每一次add时候，{} -> { {} }变成了有一个元素的集合，利用set和stack，map容器能很方便解决这道题。

            if (!mp[s1])    mp[s1] = ++cnt;
            s2.insert (mp[s1]);
        }
        sta.push (s2);
        return s2.size ();
    }
    int intersect()    {
        pop ();
        tmp.clear ();
        for (it=s1.begin (); it!=s1.end (); ++it)   {
            if (s2.find (*it) != s2.end ()) {
                tmp.insert (*it);
            }
        }
        sta.push (tmp);
        return tmp.size ();
    }
    int _union() {
        pop ();
        for (it=s1.begin (); it!=s1.end (); ++it)   {
            s2.insert (*it);
        }
        sta.push (s2);
        return s2.size ();
    }
};

int main(void)  {
    int T;  scanf ("%d", &T);
    while (T--) {
        int n;  scanf ("%d", &n);
        Opera oper;
        char str[10];
        for (int i=0; i<n; ++i) {
            scanf ("%s", &str);
            if (str[0] == 'P')  printf ("%d
", oper.push ());
            else if (str[0] == 'D') printf ("%d
", oper.dup ());
            else if (str[0] == 'A') printf ("%d
", oper.add ());
            else if (str[0] == 'U') printf ("%d
", oper._union ());
            else    printf ("%d
", oper.intersect ());
        }
        puts ("***");
    }

    return 0;
}