3 | 1月 | 2018 | 超客的水墨烟雨

uva400

maksyuki 发表于 oj 分类，标签: 基础算法-贪心, 字符串-简单字符处理

03 1月 2018

The computer company you work for is introducing a brand new computer line and is developing a new Unix-like operating system to be introduced along with the new computer. Your assignment is to write the formatter for the ls function.

Your program will eventually read input from a pipe (although for now your program will read from the input file). Input to your program will consist of a list of (F) filenames that you will sort (ascending based on the ASCII character values) and format into (C) columns based on the length (L) of the longest filename. Filenames will be between 1 and 60 (inclusive) characters in length and will be formatted into left-justified columns. The rightmost column will be the width of the longest filename and all other columns will be the width of the longest filename plus 2. There will be as many columns as will fit in 60 characters. Your program should use as few rows (R) as possible with rows being filled to capacity from left to right.

Input

The input file will contain an indefinite number of lists of filenames. Each list will begin with a line containing a single integer (1 ≤ N ≤ 100). There will then be N lines each containing one left-justified filename and the entire line’s contents (between 1 and 60 characters) are considered to be part of the filename. Allowable characters are alphanumeric (a to z, A to Z, and 0 to 9) and from the following set {._-} (not including the curly braces). There will be no illegal characters in any of the filenames and no line will be completely empty.

Immediately following the last filename will be the N for the next set or the end of file. You should read and format all sets in the input file.

Output

For each set of filenames you should print a line of exactly 60 dashes (-) followed by the formatted columns of filenames. The sorted filenames 1 to R will be listed down column 1; filenames R + 1 to 2R listed down column 2; etc.

Sample Input

10
tiny
2short4me
very_long_file_name
shorter
size-1
size2
size3
much_longer_name
12345678.123
mid_size_name
12
Weaser
Alfalfa
Stimey
Buckwheat
Porky
Joe
Darla
Cotton
Butch
Froggy
Mrs_Crabapple
P.D.
19
Mr._French
Jody
Buffy
Sissy
Keith
Danny
Lori
Chris
Shirley
Marsha
Jan
Cindy
Carol
Mike
Greg
Peter
Bobby
Alice
Ruben
Sample Output

------------------------------------------------------------
12345678.123 size-1
2short4me size2
mid_size_name size3
much_longer_name tiny
shorter very_long_file_name
------------------------------------------------------------
Alfalfa Cotton Joe Porky
Buckwheat Darla Mrs_Crabapple Stimey
Butch Froggy P.D. Weaser
------------------------------------------------------------
Alice Chris Jan Marsha Ruben
Bobby Cindy Jody Mike Shirley
Buffy Danny Keith Mr._French Sissy
Carol Greg Lori Peter

题目类型：简单字符处理

算法分析：将string排序后按照尽可能排在一行的贪心策略输出即可

/**************************************************
filename       :m.cpp
author         :maksyuki
created time   :2018/1/3 21:23:29
last modified  :2018/1/3 21:48:35
file location  :C:\Users\abcd\Desktop\TheEternalPoet
***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("in", "r", stdin)
#define CFO freopen ("out", "w", stdout)
#define CPPFF ifstream cin ("in")
#define CPPFO ofstream cout ("out")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define	DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e5 + 6666;

vector<string> ans;

int main()
{
#ifdef LOCAL
    CFF;
    //CFO;
#endif
    int n;
    while(cin >> n) {
        ans.clear();
        string s;
        int maxval = -INF;
        for(int i = 1; i <= n; i++) {
            cin >> s;
            int tmp = s.length();
            maxval = max(maxval, tmp);
            ans.push_back(s);
        }
        sort(ans.begin(), ans.end());

        int col = (60 - maxval) / (maxval + 2) + 1, row;
        if(n % col) row = n / col + 1;
        else row = n / col;

        for(int i = 1; i <= 60; i++) cout << "-";
        cout << endl;        

        string tt;
        for(int i = 0; i < row; i++) {
            for(int j = 0; j < col; j++) {
                if(i + j * row > n - 1) continue;
                tt = ans[i+j*row];
                int len = tt.length();
                cout << tt;
                int tv;
                if(j < col - 1) tv = maxval + 2;
                else tv = maxval;
                for(int k = len + 1; k <= tv; k++) cout << " ";
            }
            cout << endl;
        }
    }
    return 0;
}

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/**************************************************

filename :m.cpp

author :maksyuki

created time :2018/1/3 21:23:29

last modified :2018/1/3 21:48:35

file location :C:\Users\abcd\Desktop\TheEternalPoet

***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("in", "r", stdin)

#define CFO freopen ("out", "w", stdout)

#define CPPFF ifstream cin ("in")

#define CPPFO ofstream cout ("out")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e5 + 6666;

vector<string> ans;

int main()

{

#ifdef LOCAL

CFF;

//CFO;

#endif

int n;

while(cin >> n) {

ans.clear();

string s;

int maxval = -INF;

for(int i = 1; i <= n; i++) {

cin >> s;

int tmp = s.length();

maxval = max(maxval, tmp);

ans.push_back(s);

}

sort(ans.begin(), ans.end());

int col = (60 - maxval) / (maxval + 2) + 1, row;

if(n % col) row = n / col + 1;

else row = n / col;

for(int i = 1; i <= 60; i++) cout << "-";

cout << endl;

string tt;

for(int i = 0; i < row; i++) {

for(int j = 0; j < col; j++) {

if(i + j * row > n - 1) continue;

tt = ans[i+j*row];

int len = tt.length();

cout << tt;

int tv;

if(j < col - 1) tv = maxval + 2;

else tv = maxval;

for(int k = len + 1; k <= tv; k++) cout << " ";

}

cout << endl;

}

return 0;

}

uva540

maksyuki 发表于 oj 分类，标签: 数据结构-队列\双端\单调\优先

03 1月 2018

Queues and Priority Queues are data structures which are known to most computer scientists. The Team Queue, however, is not so well known, though it occurs often in everyday life. At lunch time the queue in front of the Mensa is a team queue, for example.

In a team queue each element belongs to a team. If an element enters the queue, it first searches the queue from head to tail to check if some of its teammates (elements of the same team) are already in the queue. If yes, it enters the queue right behind them. If not, it enters the queue at the tail and becomes the new last element (bad luck). Dequeuing is done like in normal queues: elements are processed from head to tail in the order they appear in the team queue.

Your task is to write a program that simulates such a team queue.

Input

The input file will contain one or more test cases. Each test case begins with the number of teams t (1 ≤ t ≤ 1000). Then t team descriptions follow, each one consisting of the number of elements belonging to the team and the elements themselves. Elements are integers in the range 0..999999. A team may consist of up to 1000 elements.

Finally, a list of commands follows. There are three different kinds of commands:

• ENQUEUE x — enter element x into the team queue

• DEQUEUE — process the first element and remove it from the queue

• STOP — end of test case

The input will be terminated by a value of 0 for t.

Warning: A test case may contain up to 200000 (two hundred thousand) commands, so the implementation of the team queue should be efficient: both enqueing and dequeuing of an element should only take constant time.

Output

For each test case, first print a line saying ‘Scenario #k’, where k is the number of the test case. Then, for each ‘DEQUEUE’ command, print the element which is dequeued on a single line. Print a blank line after each test case, even after the last one.

Sample Input

2
3 101 102 103
3 201 202 203
ENQUEUE 101
ENQUEUE 201
ENQUEUE 102
ENQUEUE 202
ENQUEUE 103
ENQUEUE 203
DEQUEUE
DEQUEUE
DEQUEUE
DEQUEUE
DEQUEUE
DEQUEUE
STOP
2
5 259001 259002 259003 259004 259005
6 260001 260002 260003 260004 260005 260006
ENQUEUE 259001
ENQUEUE 260001
ENQUEUE 259002
ENQUEUE 259003
ENQUEUE 259004
ENQUEUE 259005
DEQUEUE
DEQUEUE
ENQUEUE 260002
ENQUEUE 260003
DEQUEUE
DEQUEUE
DEQUEUE
DEQUEUE
STOP
0
Sample Output

Scenario #1
101
102
103
201
202
203
Scenario #2
259001
259002
259003
259004
259005
260001

题目类型：简单队列处理

算法分析：维护一个团队队列和每个组的队列，然后按照要求模拟操作即可

/**************************************************
filename       :k.cpp
author         :maksyuki
created time   :2018/1/3 20:14:39
last modified  :2018/1/3 20:28:13
file location  :C:\Users\abcd\Desktop\TheEternalPoet
***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("in", "r", stdin)
#define CFO freopen ("out", "w", stdout)
#define CPPFF ifstream cin ("in")
#define CPPFO ofstream cout ("out")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define	DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e7 + 66;
const int maxnn = 1e3 + 66;

int ma[maxn];
bool inqu[maxnn];
queue <int> qu;
queue <int> qug[maxnn];

void Init() {
    memset(inqu, false, sizeof(inqu));
    while(!qu.empty()) qu.pop();
    for(int i = 0; i < maxnn; i++) {
        while(!qug[i].empty()) qug[i].pop();
    }
}

int main()
{
#ifdef LOCAL
    CFF;
    //CFO;
#endif
    int t, id = 1;
    while(scanf("%d", &t) != EOF) {
        if(!t) break;
        
        Init();
        int ta, tb;
        for(int i = 1; i <= t; i++) {
            scanf(" %d", &ta);
            for(int j = 1; j <= ta; j++) {
                scanf("%d", &tb);
                ma[tb] = i;
            }
        }
        
        printf("Scenario #%d\n", id++);

        char s[66];
        while(scanf("%s", s) != EOF) {
            if(!strcmp(s, "STOP")) break;

            if(s[0] == 'E') {
                scanf(" %d", &ta);
                int tt = ma[ta];
                if(!inqu[tt]) {
                    qu.push(tt);
                    inqu[tt] = true;
                    qug[tt].push(ta);
                }
                else {
                    qug[tt].push(ta);
                }
            }
            else {
                int tt = qu.front();
                int tv = qug[tt].front();
                printf("%d\n", tv);
                qug[tt].pop();
                if(qug[tt].empty()) {
                    qu.pop();
                    inqu[tt] = false;
                }
            }
        }
        puts("");
    }
    return 0;
}

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/**************************************************

filename :k.cpp

author :maksyuki

created time :2018/1/3 20:14:39

last modified :2018/1/3 20:28:13

file location :C:\Users\abcd\Desktop\TheEternalPoet

***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("in", "r", stdin)

#define CFO freopen ("out", "w", stdout)

#define CPPFF ifstream cin ("in")

#define CPPFO ofstream cout ("out")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e7 + 66;

const int maxnn = 1e3 + 66;

int ma[maxn];

bool inqu[maxnn];

queue <int> qu;

queue <int> qug[maxnn];

void Init() {

memset(inqu, false, sizeof(inqu));

while(!qu.empty()) qu.pop();

for(int i = 0; i < maxnn; i++) {

while(!qug[i].empty()) qug[i].pop();

}

int main()

{

#ifdef LOCAL

CFF;

//CFO;

#endif

int t, id = 1;

while(scanf("%d", &t) != EOF) {

if(!t) break;

Init();

int ta, tb;

for(int i = 1; i <= t; i++) {

scanf(" %d", &ta);

for(int j = 1; j <= ta; j++) {

scanf("%d", &tb);

ma[tb] = i;

}

printf("Scenario #%d\n", id++);

char s[66];

while(scanf("%s", s) != EOF) {

if(!strcmp(s, "STOP")) break;

if(s[0] == 'E') {

scanf(" %d", &ta);

int tt = ma[ta];

if(!inqu[tt]) {

qu.push(tt);

inqu[tt] = true;

qug[tt].push(ta);

}

else {

qug[tt].push(ta);

}

else {

int tt = qu.front();

int tv = qug[tt].front();

printf("%d\n", tv);

qug[tt].pop();

if(qug[tt].empty()) {

qu.pop();

inqu[tt] = false;

}

puts("");

}

return 0;

}

uva12096

maksyuki 发表于 oj 分类，标签: 数据结构-栈/单调

03 1月 2018

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 in mathematics 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 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 each test 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
***

题目类型：简单数据结构

算法分析：将每个集合用int类型的ID表示，然后整个栈压入每个集合的ID，然后按照要求模拟操作即可，注意使用set_union等集合操作函数时中的sc每次都要重新定义，因为操作函数是使用inserter将数据插入到迭代器的开始端

/**************************************************
filename       :j.cpp
author         :maksyuki
created time   :2018/1/2 10:13:39
last modified  :2018/1/3 10:44:10
file location  :C:\Users\abcd\Desktop\TheEternalPoet
***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("in", "r", stdin)
#define CFO freopen ("out", "w", stdout)
#define CPPFF ifstream cin ("in")
#define CPPFO ofstream cout ("out")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define	DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e5 + 6666;

stack<int> sta;
map<int, set<int> > ma;
map<set<int>, int> mb;
int id;

void Init() {
    id = 0;
    ma.clear(), mb.clear(); 
    while(!sta.empty()) sta.pop();
}

int GetID(set<int> v) {
    if(mb.count(v)) return mb[v];
    ma[id] = v;
    return mb[v] = id++;
}

int main()
{
#ifdef LOCAL
    CFF;
    //CFO;
#endif
    
    int t;
    cin >> t;
    while(t--) {
        int n;
        cin >> n;
        string s;
        set<int> sa, sb;
        
        Init();
        for(int i = 1; i <= n; i++) {
            cin >> s;
            if(s[0] == 'P') {
                set<int> tmp;
                sta.push(GetID(tmp));            
            }
            else if(s[0] == 'D')
                sta.push(sta.top());
            else {
                sa = ma[sta.top()]; sta.pop();
                sb = ma[sta.top()]; sta.pop();
                set<int> sc;
                if(s[0] == 'U') set_union(sa.begin(), sa.end(), sb.begin(), sb.end(), inserter(sc, sc.begin()));
                if(s[0] == 'I') set_intersection(sa.begin(), sa.end(), sb.begin(), sb.end(), inserter(sc, sc.begin()));
                if(s[0] == 'A') sc = sb, sc.insert(GetID(sa));
                sta.push(GetID(sc));
            }
            cout << ma[sta.top()].size() << endl;
        }
        cout << "***" << endl;
    }
    return 0;
}

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/**************************************************

filename :j.cpp

author :maksyuki

created time :2018/1/2 10:13:39

last modified :2018/1/3 10:44:10

file location :C:\Users\abcd\Desktop\TheEternalPoet

***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("in", "r", stdin)

#define CFO freopen ("out", "w", stdout)

#define CPPFF ifstream cin ("in")

#define CPPFO ofstream cout ("out")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e5 + 6666;

stack<int> sta;

map<int, set<int> > ma;

map<set<int>, int> mb;

int id;

void Init() {

id = 0;

ma.clear(), mb.clear();

while(!sta.empty()) sta.pop();

}

int GetID(set<int> v) {

if(mb.count(v)) return mb[v];

ma[id] = v;

return mb[v] = id++;

}

int main()

{

#ifdef LOCAL

CFF;

//CFO;

#endif

int t;

cin >> t;

while(t--) {

int n;

cin >> n;

string s;

set<int> sa, sb;

Init();

for(int i = 1; i <= n; i++) {

cin >> s;

if(s[0] == 'P') {

set<int> tmp;

sta.push(GetID(tmp));

}

else if(s[0] == 'D')

sta.push(sta.top());

else {

sa = ma[sta.top()]; sta.pop();

sb = ma[sta.top()]; sta.pop();

set<int> sc;

if(s[0] == 'U') set_union(sa.begin(), sa.end(), sb.begin(), sb.end(), inserter(sc, sc.begin()));

if(s[0] == 'I') set_intersection(sa.begin(), sa.end(), sb.begin(), sb.end(), inserter(sc, sc.begin()));

if(s[0] == 'A') sc = sb, sc.insert(GetID(sa));

sta.push(GetID(sc));

}

cout << ma[sta.top()].size() << endl;

}

cout << "***" << endl;

}

return 0;

}

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