题目链接：http://codeforces.com/problemset/problem/1096/D
题意：
有一长度为n的字符串，每一字符都有一个权值，要求现在从中取出若干个字符，使得字符串中没有子串hard，（这里的子串在原串中不需要连续)。求取出字符的最小权值和。
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length nn consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 00, and removing ii-th character increases the ambiguity by aiai (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 44 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer nn (1≤n≤1051≤n≤105) — the length of the statement.
The second line contains one string ss of length nn, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤9982443531≤ai≤998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
思路：动态规划。
只要将能构成hard字符串（子串）的字符串（主串的某个子串，这个子串包含前面的那个子串）中的最后一个d之前删除同种类的字符（h，a，r，d中的某一个）使其不能构成hard。
- h的策略只能是消灭前面的h
- a的策略是将前面的h消灭，或者是以最优策略消灭前面顺序符合的
a或者ha - 同理，r的策略是使用a的策略，或者是以最优策略消灭
har或者r - d同理。
代码：
#include<stdio.h>
#include<iostream>
using namespace std;
char s[100005];
long long h,a,r,d;//(权值比较大，可能会爆范围)
long long n,x;
int main()
{
scanf("%lld",&n);
getchar();
scanf("%s",&s);
//四个策略
for(int i=0; i<n; i++)
{
scanf("%lld",&x);
if(s[i] == 'h')
h+=x;
else if(s[i]=='a')
a = min(h,a+x);
else if(s[i]=='r')
r=min(a,r+x);
else if(s[i]=='d')
d=min(r,d+x);
}
printf("%lld
",d);
return 0;
}