Codeforces 940 C. Phone Numbers

Gee_Zer

于 2019-04-19 15:13:56 发布

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分类专栏：贪心构造

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本文介绍了一种基于给定字符串s及其字母集，构造出一个长度为k的新字符串t的方法，使得t的字典序小于s，同时t的字母集为s的子集。文章详细解释了处理不同k值（k>n或k<=n）的情况，并提供了一个贪心算法实现，通过构造与s相似度高的字符串来确保t尽可能小。

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C. Phone Numbers
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
And where the are the phone numbers?

You are given a string s consisting of lowercase English letters and an integer k. Find the lexicographically smallest string t of length k, such that its set of letters is a subset of the set of letters of s and s is lexicographically smaller than t.

It’s guaranteed that the answer exists.

Note that the set of letters is a set, not a multiset. For example, the set of letters of abadaba is {a, b, d}.

String p is lexicographically smaller than string q, if p is a prefix of q, is not equal to q or there exists i, such that pi < qi and for all j < i it is satisfied that pj = qj. For example, abc is lexicographically smaller than abcd , abd is lexicographically smaller than abec, afa is not lexicographically smaller than ab and a is not lexicographically smaller than a.

Input
The first line of input contains two space separated integers n and k (1 ≤ n, k ≤ 100 000) — the length of s and the required length of t.

The second line of input contains the string s consisting of n lowercase English letters.

Output
Output the string t conforming to the requirements above.

It’s guaranteed that the answer exists.

Examples
inputCopy
3 3
abc
outputCopy
aca
inputCopy
3 2
abc
outputCopy
ac
inputCopy
3 3
ayy
outputCopy
yaa
inputCopy
2 3
ba
outputCopy
baa
Note
In the first example the list of strings t of length 3, such that the set of letters of t is a subset of letters of s is as follows: aaa, aab, aac, aba, abb, abc, aca, acb, … Among them, those are lexicographically greater than abc: aca, acb, … Out of those the lexicographically smallest is aca.

题意：给你一个序列s,用它里面的字母集合，构成一个新的序列，使它大于s，且是最小的
如例子
3 3
abc
你可以用abc这三个字母构成长度为3的序列有 aaa, aab, aac, aba, abb, abc, aca, acb, acc,…
其中大于abc的有 aca acb acc…
最小的是aca
所以输出aca

思路：
先设mn为s字符串中最小的字符，mx为是字符串中最大的字符，gz[26]为s字符串对用的集合
一共就两种情况：
第一种：如果k>n,很显然先输出s，然后再输出n-k个mn就可以了
第二种：k<=n 那么只要从k位置开始，如果它是mx，那么它肯定要改变，所以只要把末尾最大的字符串去掉，假设去掉了长度为i，那么1至i-1直接输出就可以了，接下来只要在gz[26]找到比s[i]大的第一个字符输出，然后再输出n-i-1个mn就可以了。
一个类似贪心的构造题，自己构造的字符串与s字符串相似的越多，字符串越小

#include<bits/stdc++.h>
#define LL long long
#define Max 100005
const LL mod=1e9+7;
const LL LL_MAX=9223372036854775807;
using namespace std;
int n,k;
char a[Max],gz[26];
char mn='z',mx='a';
int m[26];
int main()
{
    int len=0;
    scanf("%d%d%s",&n,&k,a);
    for(int i=0; a[i]; i++)
    {
        if(!m[a[i]-'a'])
        {
            gz[len++]=a[i];
            m[a[i]-'a']=1;
            mn=min(a[i],mn);
            mx=max(a[i],mx);
        }
    }
    sort(gz,gz+len);
    if(k>n){
       printf("%s",a);
       for(int i=n;i<k;i++)
            printf("%c",mn);
    }else{
        int i=k-1;
        while(a[i]==mx)
            i--;
       // printf("%d\n",i);
        for(int j=0;j<i;j++)
            printf("%c",a[j]);
        int j=0;
        while(a[i]>=gz[j])
            j++;
        printf("%c",gz[j]);
        for(int j=i+1;j<k;j++)
            printf("%c",mn);
    }
    return 0;
}