Mountain Subsequences（dp问题）

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本文介绍了一种算法，用于解决在给定字符序列中寻找特定山形子序列的问题。山形子序列定义为先增后减的字符序列，且长度至少为3。通过枚举中间最大值并计算左右两侧递减子序列数量，该算法能高效地找出所有符合条件的子序列，并返回其数量模2012的结果。

题目描述

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:

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
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.
The value of the letter is the ASCII value.

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

输入

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.

输出

For each case please output the number of the mountain subsequences module 2012.

样例输入

4
abca

样例输出

提示

The 4 mountain subsequences are:
aba, aca, bca, abca

思路

枚举每个字母是最中间的那个最大的字母，向两边分别找递减的子序列一共有多少个，求左右递减子序列的乘积，最终答案就是所有乘积之和

代码

#include<iostream>
#include<string>
#include<map>
#include<set>
//#include<unordered_map>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#include<iomanip>
#include<stack>
#include<cmath>
#include<fstream>
#define X first
#define Y second
#define best 131 
#define INF 0x3f3f3f3f
#define P pair<int,int>
#define ls p<<1
#define rs p<<1|1
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double eps=1e-5;
const double pai=acos(-1.0);
const int N=1e5+10;
const int maxn=1e6+10;
char s[N];
int n,dp[N],a[N],l[N],r[N],mod=2012;
int main( )
{
    int n;
    while(~scanf("%d",&n))
    {
        scanf("%s",s);
        for(int i=0;i<n;i++) a[i]=s[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]是求字母s[i]左边，结尾小于a[i]的子序列个数 
            dp[a[i]]=(dp[a[i]]+l[i]+1)%mod;
            // dp[a[i]]是求当前以a[i]为结尾的子序列的个数 
        }
        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;
            //r[i]是求字母s[i]右边，开头小于a[i]的子序列个数 
            dp[a[i]]=(dp[a[i]]+r[i]+1)%mod;// 同上 
        }
        int ans=0;
        for(int i=0;i<n;i++) ans=(ans+l[i]*r[i])%mod;
        printf("%d\n",ans);
    }
    return 0;
}