Codeforces Round #401 (Div. 2) D. Cloud of Hashtags（暴力）

D. Cloud of Hashtags
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Vasya is an administrator of a public page of organization “Mouse and keyboard” and his everyday duty is to publish news from the world of competitive programming. For each news he also creates a list of hashtags to make searching for a particular topic more comfortable. For the purpose of this problem we define hashtag as a string consisting of lowercase English letters and exactly one symbol ‘#’ located at the beginning of the string. The length of the hashtag is defined as the number of symbols in it without the symbol ‘#’.

The head administrator of the page told Vasya that hashtags should go in lexicographical order (take a look at the notes section for the definition).

Vasya is lazy so he doesn’t want to actually change the order of hashtags in already published news. Instead, he decided to delete some suffixes (consecutive characters at the end of the string) of some of the hashtags. He is allowed to delete any number of characters, even the whole string except for the symbol ‘#’. Vasya wants to pick such a way to delete suffixes that the total number of deleted symbols is minimum possible. If there are several optimal solutions, he is fine with any of them.

Input
The first line of the input contains a single integer n (1 ≤ n ≤ 500 000) — the number of hashtags being edited now.

Each of the next n lines contains exactly one hashtag of positive length.

It is guaranteed that the total length of all hashtags (i.e. the total length of the string except for characters ‘#’) won’t exceed 500 000.

Output
Print the resulting hashtags in any of the optimal solutions.

Examples
input
3
#book
#bigtown
#big
output
#b
#big
#big
input
3
#book
#cool
#cold
output
#book
#co
#cold
input
4
#car
#cart
#art
#at
output
#
#
#art
#at
input
3
#apple
#apple
#fruit
output
#apple
#apple
#fruit
Note
Word a1, a2, …, am of length m is lexicographically not greater than word b1, b2, …, bk of length k, if one of two conditions hold:

at first position i, such that ai ≠ bi, the character ai goes earlier in the alphabet than character bi, i.e. a has smaller character than b in the first position where they differ;
if there is no such position i and m ≤ k, i.e. the first word is a prefix of the second or two words are equal.
The sequence of words is said to be sorted in lexicographical order if each word (except the last one) is lexicographically not greater than the next word.

For the words consisting of lowercase English letters the lexicographical order coincides with the alphabet word order in the dictionary.

According to the above definition, if a hashtag consisting of one character ‘#’ it is lexicographically not greater than any other valid hashtag. That’s why in the third sample we can’t keep first two hashtags unchanged and shorten the other two.

题意：给出N接着是N行字符串，可以删除每个字符串的尽量短后缀，使得从上往下的字符串符合字典序相等或上升，最后一行字符串不可做任何删除。输出处理后的字符串。

思路：从最后一行开始往上遍历，删掉上一行比当前行字典序大的部分即可，毫无技巧，纯粹暴力模拟

代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<string>
#include<vector>
using namespace std;
const int maxn=500005;
vector<string>str;
int main()
{int N;scanf("%d",&N);for(int i=0; i<N; i++){string temp;cin>>temp;str.push_back(temp);}
//    for(int i=0; i<N; i++)
//        printf("%d***\n",(int)str[i].size());for(int i=N-2; i>=0; i--){if(str[i]>str[i+1])//使满足str[i]<=str[i+1];{string flag;int j=0;while(j<min((int)str[i].size(),(int)str[i+1].size())&&str[i][j]<=str[i+1][j])j++;while(j<(int)str[i].size()){str[i][j]='\0';j++;}}}for(int i=0; i<N; i++){for(int j=0; j<(int)str[i].size()&&str[i][j]!='\0'; j++)cout<<str[i][j];printf("\n");}return 0;
}

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