$\color{green}{UVa12096}$

#include<iostream>
#include<stack>
#include<set>
#include<cstring>
#include<map>
#include<vector>
#include<algorithm>
#include <iterator>
using namespace std;

typedef set<int>Set;
//给set<int>起个别名叫Set


stack<int> s;
map<Set, int> IDcache;
vector<Set> Setcache;

int ID(Set x) {//这个函数的参数是 整型集合
    if (IDcache.count(x)) //如果存在返回1，否则返回0
        return IDcache[x];//返回集合的编号
    Setcache.push_back(x);//否则添加新集合
    return IDcache[x] = Setcache.size() - 1;//分配编号，是第几个集合就分配第几
}

int main()
{
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        while (n--) {
            string op;
            cin >> op;
            if (op[0] == 'P') s.push(ID(Set()));//实例化一个空的set<int>传给函数ID
            else if (op[0] == 'D') s.push(s.top());
            else {
                Set x1 = Setcache[s.top()];s.pop();
                Set x2 = Setcache[s.top()];s.pop();
                Set x;
                if (op[0] == 'U') set_union(x1.begin(), x1.end(), x2.begin(), x2.end(), inserter(x,x.begin()));
                if (op[0] == 'I') set_intersection(x1.begin(), x1.end(), x2.begin(), x2.end(), inserter(x, x.begin()));
                if (op[0] == 'A') { x = x2;x.insert(ID(x1)); }
                s.push(ID(x));
            }
            cout << Setcache[s.top()].size() << endl;
        }
        cout << "***" << endl;
    }
    return 0;
}

[√] UVA540 团体队列 Team Queue

[√] UVA136 丑数 Ugly Numbers

数学题……嗯，题目意思不是三个因子都含有的数才叫丑数，是至少含有其中之一就可以了

然后丑数的2倍、3倍、5倍都是丑数，注意这里可能出现重复，紫书上用set就是为了去重

这里优先队列的话就记录一下队首元素的值，多次pop()直到下一个不同丑数为止

$\color{green}{UVa136}$

#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<iterator>
#include<set>
#include<queue>
using namespace std;
typedef long long ll;
int main()
{
  priority_queue<ll, vector<ll>, greater<ll> > q;
  q.push(1);
  for (int i = 1;;i++) {
      ll w = q.top();
      if (i == 1500) {
          cout << "The 1500'th ugly number is " << w<< "."<<endl;
          return 0;
      }
      q.push(w * 2);
      q.push(w * 3);
      q.push(w * 5);
      while (w == q.top()) q.pop();
  }
  return 0;
}

UVA400 Unix Is命令 Unix ls

[√] UVA1592 数据库 Database

利用map.count()来判断是否已经出现过，如此就仅需要遍历一次n（行）。

$\color{green}{UVa1592}$

#include<iostream>
#include<stack>
#include<set>
#include<cstring>
#include<map>
#include<vector>
#include<algorithm>
#include <iterator>
using namespace std;

//ID将字符串对应到唯一的编号，将整个表格v数字化
//mp将编号的组合对应到列的组合

long long v[10000 + 10][15];
int n, m;
string s1;
int main() {
    while (cin >> n >> m) {
        int id = 0;
        map<string, int> ID;
        map<long long, int> mp;
        getchar();
        memset(v, 0, sizeof(v));
        int i, j;
        for (i = 1;i <= n;i++) {
            for (j = 1;j <= m;j++) {
                s1.clear();
                while (1) {
                    char ch = getchar();
                    if (ch == ',' || ch == '\n' || ch == '\r' || ch == EOF) break;
                    s1.push_back(ch);
                }
                if (ID.count(s1)) v[i][j] = ID[s1];
                else v[i][j] = ID[s1] = ++id;
            }
        }
        int ok = 0;
        for (i = 1;i <= m;i++) {
            for (j = i + 1;j <= m;j++) {
                mp.clear();
                for (int k = 1;k <= n;k++) {
                    if (mp.count((long long)v[k][i] * 1000000 + v[k][j])) {
                        cout << "NO" << endl;
                        cout << mp[(long long)v[k][i] * 1000000 + v[k][j]] << " " << k << endl;
                        cout << i << " " << j << endl;
                        ok = 1;
                    }
                    else mp[(long long)v[k][i] * 1000000 + v[k][j]] = k;
                    if (ok) break;
                }
            }
        }
        if (!ok) cout << "YES" << endl;
    }
    return 0;
}