zoukankan      html  css  js  c++  java
  • Don't Be a Subsequence

    问题 F: Don't Be a Subsequence

    时间限制: 1 Sec  内存限制: 128 MB
    提交: 33  解决: 2
    [提交] [状态] [讨论版] [命题人:]

    题目描述

    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.

    输入

    Input is given from Standard Input in the following format:
    A

    输出

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

    样例输入

    atcoderregularcontest
    

    样例输出

    b
    

    提示

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


    分析:这题。。勉强算个图论吧

    • 首先,动态规划确定最短子串的长度,dp[i]为[i….)这一段中最短的非子序列长度。
    • 则dp[i]=min(dp[i],dp[next[i][j]+1]+1)
    • 再反着走回来枚举26个字母,取字典序最小的字母输出。
    #include <iostream>
    #include <string>
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <deque>
    #include <map>
    #define range(i,a,b) for(auto i=a;i<=b;++i)
    #define LL long long
    #define itrange(i,a,b) for(auto i=a;i!=b;++i)
    #define rerange(i,a,b) for(auto i=a;i>=b;--i)
    #define fill(arr,tmp) memset(arr,tmp,sizeof(arr))
    using namespace std;
    int dp[int(2e5+5)],alpha[30],NEXT[int(2e5+5)][30],len;
    string word;
    void init(){
        cin>>word;
        len=int(word.size());
        range(i,0,25)alpha[i]=len;
        rerange(i,len-1,0){
            alpha[word[i]-'a']=i;
            range(j,0,25)NEXT[i][j]=alpha[j];
            dp[i]=int(1e9+7);
        }
        dp[len]=1;
    }
    void solve(){
        rerange(i,len-1,0)
        range(j,0,25)dp[i]=min(dp[i],dp[NEXT[i][j]+1]+1);
        int pos=0;
        rerange(i,dp[0],0)range(j,0,25)if(dp[pos]==dp[NEXT[pos][j]+1]+1){
            putchar('a'+j);
            pos=NEXT[pos][j]+1;
            break;
        }
    }
    int main() {
        init();
        solve();
        return 0;
    }
    View Code
  • 相关阅读:
    模拟登陆+数据爬取 (python+selenuim)
    matplotlib基本使用(矩形图、饼图、热力图、3D图)
    tensorflow进阶篇-4(损失函数1)
    CS231n学习笔记-图像分类笔记(下篇)
    CS231n学习笔记-图像分类笔记(上篇)
    numpy 基本使用1
    tensorflow基础篇-2
    tensorflow进阶篇-3
    tensorflow基础篇-1
    自定义滚动条第一版
  • 原文地址:https://www.cnblogs.com/Rhythm-/p/9377038.html
Copyright © 2011-2022 走看看