D. Cloud of Hashtags

题目链接

题意：
\(n\)个以\(#\)开头的由小写字母组成的字符串，求删除最少字母数使得字符串\(s_i\)的字典序不大于\(s_{i+1}\).

思路：
从前往后不好比较，逆向思维。最后一个字符串字典序肯定尽可能大，即不删除字母，然后倒着模拟一下即可。

code:

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <deque>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#include <unordered_map>

#define fi first
#define se second
#define pb push_back
#define endl "\n"
#define debug(x) cout << #x << ":" << x << endl;
#define bug cout << "********" << endl;
#define all(x) x.begin(), x.end()
#define lowbit(x) x & -x
#define fin(x) freopen(x, "r", stdin)
#define fout(x) freopen(x, "w", stdout)
#define ull unsigned long long
#define ll long long 

const double eps = 1e-15;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
const int mod =  1e9 + 7;
const int maxn = 5e5 + 10;

using namespace std;

string s[maxn];
int n;

int main(){
    ios::sync_with_stdio(false), cin.tie(0);
    cin >> n;
    for(int i = 1; i <= n; i ++)cin >> s[i];

    for(int i = n - 1; i >= 1; i --){
        int len = s[i].size();
        for(int j = 0; j < len; j ++){
            if(s[i][j] < s[i + 1][j])break;
            else if(s[i][j] > s[i + 1][j] || j >= s[i].size()){
                s[i].erase(j);
                break;
            }
        }
    }

    for(int i = 1; i <= n; i ++)cout << s[i] << endl;
    return 0;
}