CF - 797C. Minimal string - 贪心+栈处理+字符串处理

• 题目描述：
  C. Minimal string

  time limit per test

  1 second

  memory limit per test

  256 megabytes

  input

  standard input

  output

  standard output

  Petya recieved a gift of a string s with length up to 105 characters for his birthday. He took two more empty strings t and u and decided to play a game. This game has two possible moves:

  • Extract the first character of s and append t with this character.
  • Extract the last character of t and append u with this character.

  Petya wants to get strings s and t empty and string u lexigraphically minimal.

  You should write a program that will help Petya win the game.

  Input

  First line contains non-empty string s (1 ≤ |s| ≤ 105), consisting of lowercase English letters.

  Output

  Print resulting string u.

  Examples

  input

  cab
  

  output

  abc
  

  input

  acdb
  

  output

  abdc

• 题意概述：给出string s长度<=1e5, op1:把s的第一个字符移动到t末尾.op2:把t最后一个字符移到u末尾,求u能得到的最小字典序?
• 解题思路：先预处理每个字符个数，那么在输出第i位时候，先把它压人栈内，判断栈顶的字符是不是当前状态下字典序最小的字符，如果是则把栈内全部字符输出，不是则继续。
• AC代码：

  #include <bits/stdc++.h>
  #define INF 0x3f3f3f3f
  #define maxn 100100
  #define lson root << 1
  #define rson root << 1 | 1
  #define lent (t[root].r - t[root].l + 1)
  #define lenl (t[lson].r - t[lson].l + 1)
  #define lenr (t[rson].r - t[rson].l + 1)
  #define N 1111
  #define eps 1e-6
  #define pi acos(-1.0)
  #define e exp(1.0)
  using namespace std;
  const int mod = 1e9 + 7;
  typedef long long ll;
  typedef unsigned long long ull;
  int vis[26];
  string str;
  stack<char> s;
  vector<int> ans;
  int main()
  {
  #ifndef ONLINE_JUDGE
  	freopen("in.txt", "r", stdin);
  	freopen("out.txt", "w", stdout);
  	long _begin_time = clock();
  #endif
  	while (cin >> str)
  	{
  		memset(vis, 0, sizeof(vis));
  		ans.clear();
  		while (!s.empty())
  			s.pop();
  		int len = str.length();
  		for (int i = 0; i < len; i++)
  			vis[str[i] - 'a']++;
  		for (int i = 0; i < len; i++)
  		{
  			int pos = str[i] - 'a';
  			s.push(pos);
  			vis[pos]--;
  			bool flag = 1;
  			while (!s.empty())
  			{
  				int cur = s.top();
  				for (int i = 0; i < cur; i++)
  					if (vis[i])
  					{
  						flag = 0;
  						break;
  					}
  				if (!flag)
  					break;
  				ans.push_back(cur);
  				s.pop();
  			}
  		}
  		int sz = ans.size();
  		for (int i = 0; i < sz; i++)
  			printf("%c", ans[i] + 'a');
  		puts("");
  	}
  #ifndef ONLINE_JUDGE
  	long _end_time = clock();
  	printf("time = %ld ms.", _end_time - _begin_time);
  #endif
  	return 0;
  }