Mountain Subsequences(dp)

本文介绍了一种高效算法,用于计算给定字符串中符合特定条件的山形子序列的数量。该算法通过两次遍历字符串来计算每个字符左侧的递增子序列数和右侧的递减子序列数。

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思路:

遍历两次求得每个字符左侧的递增子序列数和右侧的递减子序列数。

最后再遍历一遍每个字符左右序列数相乘。


代码:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1e5+5;
const int mod = 2012;
char str[maxn];
int n, a[maxn], l[maxn], r[maxn], dp[maxn];

int main(void)
{
    while(cin >> n)
    {
        scanf(" %s", str);
        for(int i = 0; i < n; i++)
            a[i] = str[i]-'a';
        memset(l, 0, sizeof(l));
        memset(r, 0, sizeof(r));
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < a[i]; j++)
                l[i] = (l[i]+dp[j])%mod;    //l[i]统计前i个能组成结尾小于a[i]的方案数
            dp[a[i]] = (dp[a[i]]+l[i]+1)%mod;   //dp[a[i]]统计当前位置以a[i]字符结尾的方案数
        }
        memset(dp, 0, sizeof(dp));
        for(int i = n-1; i >= 0; i--)
        {
            for(int j = 0; j < a[i]; j++)
                r[i] = (r[i]+dp[j])%mod;
            dp[a[i]] = (dp[a[i]]+r[i]+1)%mod;
        }
        int ans = 0;
        for(int i = 0; i < n; i++)
            ans = (ans+l[i]*r[i])%mod;
        printf("%d\n", ans);
    }
    return 0;
}


Mountain Subsequences

Time Limit: 1000MS Memory Limit: 65536KB
Problem Description

Coco is a beautiful ACMer girl living in a very beautiful mountain. There are many trees and flowers on the mountain, and there are many animals and birds also. Coco like the mountain so much that she now name some letter sequences as Mountain Subsequences.

 

A Mountain Subsequence is defined as following:

1. If the length of the subsequence is n, there should be a max value letter, and the subsequence should like this, a1 < ...< ai < ai+1 < Amax > aj > aj+1 > ... > an

2. It should have at least 3 elements, and in the left of the max value letter there should have at least one element, the same as in the right.

3. The value of the letter is the ASCII value.

 

Given a letter sequence, Coco wants to know how many Mountain Subsequences exist.

Input
Input contains multiple test cases.

For each case there is a number n (1<= n <= 100000) which means the length of the letter sequence in the first line, and the next line contains the letter sequence.

Please note that the letter sequence only contain lowercase letters. 

Output
For each case please output the number of the mountain subsequences module 2012.
Example Input
4abca
Example Output
4
Hint
The 4 mountain subsequences are:

aba, aca, bca, abca


### 关于回文子序列的算法及其示例 #### 定义与概念 回文是指正读和反读都相同的字符序列。对于给定字符串中的任意字符组合形成的子序列,如果该子序列满足上述条件,则称为回文子序列。 #### 动态规划求解最长回文子序列 为了找到一个字符串中最长的回文子序列,可以采用动态规划的方法来解决这个问题。设 `dp[i][j]` 表示从第 i 到 j 的子串内的最长回文子序列长度: - 当 s[i]==s[j] 时, dp[i][j]=dp[i+1][j−1]+2; - 否则, dp[i][j]=max(dp[i+1][j],dp[i][j−1]). 最终的结果保存在 `dp[0][len(s)-1]` 中[^3]. ```python def longest_palindromic_subseq(s: str) -> int: n = len(s) # 创建二维数组用于存储中间结果 dp = [[0]*n for _ in range(n)] # 初始化单个字符的情况 for i in range(n): dp[i][i] = 1 # 填充表格 for length in range(2, n + 1): for start in range(n - length + 1): end = start + length - 1 if s[start] == s[end]: dp[start][end] = dp[start+1][end-1] + 2 else: dp[start][end] = max(dp[start+1][end], dp[start][end-1]) return dp[0][-1] ``` 此方法的时间复杂度为 O(),空间复杂度同样为 O(). #### 枚举所有可能的回文子序列 除了寻找最长的回文子序列外,还可以通过枚举的方式找出所有的不同回文子序列。这种方法适用于较短的输入字符串,并且可以通过位掩码技术实现高效的遍历。 ```python from collections import defaultdict def count_distinct_palindrome_subsequences(text: str) -> list[str]: results = set() memo = {} def backtrack(start=0, path=""): nonlocal text, results, memo key = (start, path) if key not in memo: temp_set = {path} if path == path[::-1] else {} for index in range(start, len(text)): new_path = path + text[index] if new_path == new_path[::-1]: temp_set.add(new_path) temp_set |= backtrack(index + 1, new_path) memo[key] = temp_set results.update(memo[(start, path)]) return memo[(start, path)] backtrack() return sorted(list(results)) ``` 这段代码会返回按字典序排列的不同回文子序列列表.
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