AtCoder Regular Contest 081 E

E - Don't Be a Subsequence

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Time limit : 2sec / Memory limit : 256MB

Score : 600 points

Problem Statement

A subsequence of a string S is a string that can be obtained by deleting zero or more characters from S without changing the order of the remaining characters. For example, arc, artistic and (an empty string) are all subsequences of artistic; abc and ci are not.

You are given a string A consisting of lowercase English letters. Find the shortest string among the strings consisting of lowercase English letters that are not subsequences of A. If there are more than one such string, find the lexicographically smallest one among them.

Constraints

• 1≤|A|≤2×105
• A consists of lowercase English letters.

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Input

Input is given from Standard Input in the following format:

A

Output

Print the lexicographically smallest string among the shortest strings consisting of lowercase English letters that are not subsequences of A.

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Sample Input 1

Copy

atcoderregularcontest

Sample Output 1

Copy

b

The string atcoderregularcontest contains a as a subsequence, but not b.

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Sample Input 2

Copy

abcdefghijklmnopqrstuvwxyz

Sample Output 2

Copy

aa

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Sample Input 3

Copy

frqnvhydscshfcgdemurlfrutcpzhopfotpifgepnqjxupnskapziurswqazdwnwbgdhyktfyhqqxpoidfhjdakoxraiedxskywuepzfniuyskxiyjpjlxuqnfgmnjcvtlpnclfkpervxmdbvrbrdn

Sample Output 3

Copy

aca


dp[i]代表[i~n]的子串的最小字典序串的长度

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#define ll long long
#define pb push_back
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
#define cls(name,x) memset(name,x,sizeof(name))
#define fs first
#define sc second
#define mp make_pair
#define L(x) (1<<x)
#define next Next
using namespace std;
const int inf=1e9+10;
const ll llinf=1e16+10;
const int maxn=2e5+10;
const int maxm=1e3+10;
const int mod=1e9+7;
int n;
vector<int> v[30];
char s[maxn];
int dp[maxn];
int nch=26;
int querry(int pos,int ch)
{
    vector<int>::iterator it;
    it=lower_bound(v[ch].begin(),v[ch].end(),pos);
    if(it==v[ch].end()) return n;
    else return *it;
}
void solve()
{
    dp[n+1]=0;
    for(int i=n-1;i>=0;i--)
    {
        for(int j=0;j<nch;j++)
            dp[i]=min(dp[i],dp[querry(i,j)+1]+1);
    }
    int pos=0;
    while(pos<n)
    {
        for(int j=0;j<nch;j++)
        {
            int t=querry(pos,j)+1;
            if(dp[pos]==dp[t]+1)
            {
                printf("%c",j+'a');
                pos=t;
                break;
            }
        }
    }
    printf("
");
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%s",s))
    {
        for(int i=0;i<nch;i++)
            v[i].clear();
        n=strlen(s);
        for(int i=0;i<=n;i++)
            dp[i]=inf;
        for(int i=0;i<n;i++)
            v[s[i]-'a'].pb(i);
        solve();
    }
    return 0;
}