【Codeforces 279C】Ladder

最新推荐文章于 2019-09-04 22:12:46 发布

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最新推荐文章于 2019-09-04 22:12:46 发布

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原文链接：http://www.cnblogs.com/AWCXV/p/10586324.html

本文介绍了一种通过预处理前后递增子序列长度的方法来快速判断任意两个位置间是否存在递增子序列的问题解决思路。利用这种方法可以高效地解决特定条件下的递增子序列判断问题。

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【链接】我是链接,点我呀:)
【题意】

题意

【题解】

设pre[i]表示i往前一直递增能递增多远
设aft[i]表示i往后一直递增能递增多远
如果aft[l]+pre[r]>=(r-l+1)那么就ok否则no

【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 = 50000;
    static class Task{
        
        int n,m;
        
        public void solve(InputReader in,PrintWriter out) {
            n = in.nextInt();m = in.nextInt();
            int a[] = new int[n+5];
            int aft[] = new int[n+5],pre[] = new int[n+5];
            for (int i = 1;i <= n;i++) a[i] = in.nextInt();
            for (int i = n;i >= 1;i--) {
                if (i+1<=n && a[i]<=a[i+1]) 
                    aft[i] = aft[i+1]+1;
                else
                    aft[i] = 1;
            }
            
            for (int i = 1;i <= n;i++) {
                if (i-1>=1 && a[i-1]>=a[i])
                    pre[i] = pre[i-1]+1;
                else
                    pre[i] = 1;
            }
            for (int i = 1;i <= m;i++) {
                int x,y;
                x = in.nextInt();y = in.nextInt();
                if (aft[x]+pre[y]>=(y-x+1)) {
                    System.out.println("Yes");
                }else {
                    System.out.println("No");
                }
            }
        }
    }

    

    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());
        }
    }
}

转载于:https://www.cnblogs.com/AWCXV/p/10586324.html