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最新推荐文章于 2022-02-19 11:52:10 发布

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#二分 #贪心

本文解析了CodeForces竞赛中一道关于字符串分割的问题。通过分析题目要求，给出了使用二分查找来寻找最优字符串宽度的方法，并附带完整代码实现。

The main city magazine offers its readers an opportunity to publish their ads. The format of the ad should be like this:

There are space-separated non-empty words of lowercase and uppercase Latin letters.

There are hyphen characters '-' in some words, their positions set word wrapping points. Word can include more than one hyphen.

It is guaranteed that there are no adjacent spaces and no adjacent hyphens. No hyphen is adjacent to space. There are no spaces and no hyphens before the first word and after the last word.

When the word is wrapped, the part of the word before hyphen and the hyphen itself stay on current line and the next part of the word is put on the next line. You can also put line break between two words, in that case the space stays on current line. Check notes for better understanding.

The ad can occupy no more that k lines and should have minimal width. The width of the ad is the maximal length of string (letters, spaces and hyphens are counted) in it.

You should write a program that will find minimal width of the ad.

Input

The first line contains number k (1 ≤ k ≤ 10⁵).

The second line contains the text of the ad — non-empty space-separated words of lowercase and uppercase Latin letters and hyphens. Total length of the ad don't exceed 10⁶ characters.

Output

Output minimal width of the ad.

Examples

Input

Copy

4
garage for sa-le

Output

Copy

Input

Copy

4
Edu-ca-tion-al Ro-unds are so fun

Output

Copy

Note

Here all spaces are replaced with dots.

In the first example one of possible results after all word wraps looks like this:

garage.
for.
sa-
le

The second example:

Edu-ca-
tion-al.
Ro-unds.
are.so.fun

题目大意:

给你一个k，一列字符串。这个字符串中,空格和连字符-都是标志一个单词的结束。然后让你将这个字符串分成最多k块。使得长度最长的内个块的长度尽可能的小。分块时，只能在空格处和连字符处断开.

思路:

一个典型的最大值最小值问题。设长度最长的单词长度为maxn，字符串长度为num.则答案必定在maxn与num之间。那么就二分答案。每次判断mid,此时的mid就是块划分的依据，用sum累加，当加上下一个单词大于mid时，就产生了一个块.最后判断这个块是否成立，成立说明这个划分依据大了，那就r=mid-1，继续判断.最后输出l，就是最佳划分

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#define ll long long
using namespace std;
const int inf=0x3f3f3f3f;
const int N =1e6+5;
char a[N];
int loc[N];
int cnt;
bool judge(int mid,int k)
{
    int cc=0;
    ll sum=0;
    for(int i=0;i<cnt;i++)
    {
        sum+=loc[i];
        if(sum>mid)//每当超过mid，那就产生一个块，块数量cc加1
        {
            cc++;
            sum=loc[i];
            if(cc>=k)//要注意最后的cc如果等于k，那就代表着此时有cc+1个划分快，不满足条件
            {
                return false;
            }
        }
    }
    return true;
}
int main()
{
    int k;
    cin>>k;
    getchar();
    gets(a);
    int num=strlen(a);
    //cout<<num<<endl;
    int maxn=1;
    cnt=0;
    int sign=-1;
    for(int i=0;i<num;i++)
    {
        if(a[i]==' '||a[i]=='-')
        {
           loc[cnt++]=i-sign;//先预处理出每个单词的长度，注意这个长度是带上空格或者连字符的
           maxn=max(maxn,i-sign);
           sign=i;
        }
    }
    loc[cnt++]=num-sign-1;
    maxn=max(maxn,num-sign-1);
    int l=maxn,r=num;//答案必定在maxn与num之间。
    while(l<=r)//开始二分答案
    {
       //cout<<l<<' '<<r<<endl;
        int mid=(l+r)>>1;
        if(judge(mid,k))
            r=mid-1;
        else
            l=mid+1;
    }
    cout<<l<<endl;
    return 0;
}