【Codeforces 797 C. Minimal string 】队列+栈

CF 797C
• 给定长度为n的小写字母字符串s，及空串t, u，两种操作
• 1. 将s的第一个字符加到t的末尾
• 2. 将t的最后一个字符加到u的末尾
• 求字典序最小的字符串u (长度必须为n，即s, t最后为空串)
• 1 ≤ n ≤ 1e5

这道题骚在 s 是一个队列
t 是一个栈
那么只要你s中最小的字符比栈的元素小于
那么肯定是队列进栈
否则栈去 t 中

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/

const int MAX_N = 100025;
char str[MAX_N];
queue<char > q;
int cnt[30];
stack<char >st;
char minn()
{
    for(int i = 1;i<=26;++i)
    {
        if(cnt[i])
        {
            return 'a'-1+i;
        }
    }
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    scanf("%s",str+1);
    string ans = "";
    int len = strlen(str+1);
    for(int i = 1;i<=len;++i)
    {
        q.push(str[i]);
        cnt[str[i]-'a'+1]++;
    }
    while(!q.empty())
    {
        char minn_ = minn();
        if(st.empty())
        {
            char now = q.front();
            cnt[now-'a'+1]--;
            q.pop();
            st.push(now);
        }
        else
        {
            if(st.top()<=minn_)
            {
                char top = st.top();
                st.pop();
                ans+=top;
            }
            else
            {
                char now = q.front();
                cnt[now-'a'+1]--;
                q.pop();
                st.push(now);
            }
        }
    }
    while(!st.empty())
    {
        ans+=st.top();
        st.pop();
    }
    printf("%s",ans.c_str());
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}