【鏈接】 我是鏈接,點我呀:)
【題意】

讓你將一個字符串刪掉一些字符。
使得字符串中不包含子序列"hard"
刪掉每個字符的代價已知為ai
讓你求出代價最小的方法.

【題解】

設dp[i][j]表示前i個字符,已經和"hard"匹配前j個的最小花費。
對於dp[i][j]
對s[i+1]分類討論
①s[i+1]不刪
那麽有兩種情況
第一種:s[i+1]和"hard"的第j+1個字符匹配
第二種:..xxxxx不匹配
則分別轉移到dp[i+1][j+1]和dp[i+1][j]
②s[i+1]刪掉
轉移到dp[I+1][j]且用dp[i][j]+a[i+1]嘗試轉移。

【代碼】

import java.io.*;
import java.util.*;

public class Main {
    
    
    static InputReader in;
    static PrintWriter out;
        
    public static void main(String[] args) throws IOException{
        //InputStream ins = new FileInputStream("E:\\rush.txt");
        InputStream ins = System.in;
        in = new InputReader(ins);
        out = new PrintWriter(System.out);
        //code start from here
        new Task().solve(in, out);
        out.close();
    }
    
    static int N = (int)1e5;
    static class Task{
        
        int n;
        String s;
        long a[] = new long[N+10];
        String goal=new String(" hard");
        long dp[][] = new long[N+10][10];
        
        public void solve(InputReader in,PrintWriter out) {
            n = in.nextInt();
            s = in.next();
            s = " "+s;
            for (int i = 1;i <=n;i++) a[i] = in.nextLong();
            for (int i = 0;i <= N;i++) 
                for (int j = 0;j <= 8;j++)
                    dp[i][j] = (long)(1e17);
            dp[0][0] = 0;
            for(int i = 0;i < n;i++)
                for (int j = 0;j <= 3;j++) {
                    //第i+1個不刪
                    if (s.charAt(i+1)==goal.charAt(j+1)) {
                        dp[i+1][j+1] = Math.min(dp[i+1][j+1], dp[i][j]);
                    }else {
                        dp[i+1][j] = Math.min(dp[i+1][j], dp[i][j]);
                    }
                    
                    //第i+1個刪掉
                    dp[i+1][j] = Math.min(dp[i+1][j], dp[i][j]+a[i+1]);
                }
            long ans = dp[n][0];
            for (int i = 1;i <= 3;i++) {
                ans = Math.min(ans, dp[n][i]);
            }
            out.println(ans);
        }
    }

    

    static class InputReader{
        public BufferedReader br;
        public StringTokenizer tokenizer;
        
        public InputReader(InputStream ins) {
            br = new BufferedReader(new InputStreamReader(ins));
            tokenizer = null;
        }
        
        public String next(){
            while (tokenizer==null || !tokenizer.hasMoreTokens()) {
                try {
                tokenizer = new StringTokenizer(br.readLine());
                }catch(IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(next());
        }
        
        public long nextLong() {
            return Long.parseLong(next());
        }
    }
}
 

【Codeforces 1096D】Easy Problem