第四届 Mountain Subsequences

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.
Sample Input
4abca
Sample Output
4
Hint
The 4 mountain subsequences are:     aba, aca, bca, abca

题目大意：
给你一个长度为n的字符串且仅由小写英文字母组成，求满足
a1 < ...< ai < ai+1 < Amax > aj > aj+1 > ... > an
的子序列的个数,其实也就是统计所有满足以某一元素为中心左边递增，右边递减的子序列的数目，要求该子序列最小长度为3，中心元素左右都至少有一个元素。
思路解析：
题目告诉我们输入串仅包含26个英文字母，这样我们只要枚举每一个位置，然后记录每个位置左边满足条件的个数，右边满足条件的个数，最后乘起来就是了。关键是我们如何统计左右两边满足的个数呢？因为有之前做过的求最长递增子序列的基础，所以可以应用DP的阶段划分的思想。那么，怎么划分阶段呢？我们定义数组dp[26], dl[maxn], dr[maxn],  dl[i]表示第i个位置左边满足条件的个数，dr[i]表示第i个位置右边满足条件的个数，dp[c]表示以字符c结尾的满足情况的个数。我们可以发现，dp[s[i]] = (dl[i] + 1),s[i]= c; 
1表示s[i]单独一个时满足的情况，dl[i]表示他左边的满足的各种情况加上s[i] 后满足的情况。
注意：在一个串中如果有相同的字母，那么他们的dp值是否相同呢？题目在要求该子串的时候用了
a1 < ...< ai < ai+1 < Amax > aj > aj+1 > ... > an 这样一个式子，每个元素都表示不同位置的元素，也就是说，在串
aaba中应该有aba，aba两个答案，
#include <iostream>
#include <string.h>
using namespace std;
const int mod=2012;
int dp[30],dl[100005],dr[100005],s[100005];
int main()
{
    int n;
    while(cin>>n)
    {
        string str;
        cin>>str;
        memset(dp,0,sizeof(dp));
        memset(dl,0,sizeof(dl));
        memset(dr,0,sizeof(dr));
        for(int i=0;i<n;i++)
            s[i]=str[i]-'a';
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<s[i];j++)
                dl[i]=(dl[i]+dp[j])%mod;
            dp[s[i]]=(dp[s[i]]+dl[i]+1)%mod;
        }
        memset(dp,0,sizeof(dp));
        for(int i=n-1;i>=0;i--)
        {
            for(int j=0;j<s[i];j++)
                dr[i]=(dr[i]+dp[j])%mod;
            dp[s[i]]=(dp[s[i]]+dr[i]+1)%mod;
        }
        int ans=0;
        for(int i=0;i<n;i++)
            ans=(ans+dl[i]*dr[i])%mod;
        cout<<ans<<endl;
    }
    return 0;
}