Codeforces 797C Minimal string【贪心】

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

题目大意：

给你一个栈，然你找到一个出栈顺序，使得字典序最小。

思路：

逆序维护一个数组minn【i】=x，表示第i个位子后边最小的字符是x.

那么对应维护一个栈，如果此时栈顶字符小于等于minn【此时要加入的元素的位子】，那么就出栈，将栈顶这个字符输出；

同时每个字符都在操作结束后入栈。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<stack>
using namespace std;
char ss[100800];
int a[100800];
int minn[100800];
int main()
{
while(~scanf("%s",ss))
{
int n=strlen(ss);
memset(minn,0,sizeof(minn));
for(int i=0;i<n;i++)a[i]=ss[i]-'a'+1;
for(int i=n-1;i>=0;i--)
{
if(i==n-1)minn[i]=a[i];
else minn[i]=min(minn[i+1],a[i]);
}
stack<char >s;
for(int i=0;i<n;i++)
{
if(s.size()==0)
{
s.push(ss[i]);
}
else
{
while(!s.empty())
{
int u=s.top()-'a'+1;
if(u<=minn[i])
{
printf("%c",s.top());
s.pop();
}
else break;
}
s.push(ss[i]);
}
}
while(!s.empty())
{
printf("%c",s.top());
s.pop();
}
printf("\n");
}
}