【Codeforces 1096D】Easy Problem

【链接】 我是链接,点我呀:)

【题意】

让你将一个字符串删掉一些字符。
使得字符串中不包含子序列"hard"
删掉每个字符的代价已知为ai
让你求出代价最小的方法.

【题解】

设dp[i][j]表示前i个字符,已经和"hard"匹配前j个的最小花费。
对于dp[i][j]
对s[i+1]分类讨论
①s[i+1]不删
那么有两种情况
第一种:s[i+1]和"hard"的第j+1个字符匹配
第二种:..xxxxx不匹配
则分别转移到dp[i+1][j+1]和dp[i+1][j]
②s[i+1]删掉
转移到dp[I+1][j]且用dp[i][j]+a[i+1]尝试转移。

【代码】

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个不删
        			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个删掉
        			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());
        }
    }
}