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.
First line contains non-empty string s (1 ≤ |s| ≤ 105), consisting of lowercase English letters.
Print resulting string u.
cab
abc
acdb
abdc
題目大意:給一個字符串s,兩個空字符串t,u,有兩種操作,1.將s的第一個字母加到t後,2.將t的最後一個字母加到u後,最後得到字典序最小的字符串u
思路:統計s中每個字母出現的次數,用棧模擬t,第一個字母一定是s中最小的字母,之後有兩種選擇,如果當前沒入棧的字母中有比棧頂字母小的,則繼續入棧,若果沒有,則出棧並將棧頂字母給u,每次判斷棧頂元素後有無比它小的字母
#include <bits/stdc++.h>
#define manx 100005
using namespace std;
int a[27];
bool found(char c) //當前字符的後面是否有比它小的字符,有則s to t,否則 t to u
{
int n=c-'a';
for (int i=0; i<n; i++){
if (a[i]) return true;
}
return false;
}
int main()
{
char s[manx],u[manx];
while(~scanf("%s",s)){
stack<char>q;
memset(a,0,sizeof(a));
memset(u,0,sizeof(u));
int l=strlen(s);
for (int i=0; i<l; i++) //統計每個字母出現的次數
a[s[i]-'a']++;
int j=0;
q.push(s[0]);
a[s[0]-'a']--;
int cot=0;
for (int i=1; i<l; i++){
while (!q.empty()&& !found(q.top()) ){
char t=q.top();
u[cot++]=t;
q.pop();
}
q.push(s[i]);
a[s[i]-'a']--; //a表示沒入棧的每個字母的個數
}
while(!q.empty()){
u[cot++]=q.top();
q.pop();
}
puts(u);
}
return 0;
}