• hdu 5532(最长上升子序列)


    Input
    The first line contains an integer T indicating the total number of test cases. Each test case starts with an integer n in one line, then one line with n integers a1,a2,,an.

    1T2000
    2n105
    1ai105
    There are at most 20 test cases with n>1000.
     
    Output
    For each test case, please output "`YES`" if it is almost sorted. Otherwise, output "`NO`" (both without quotes).
     
    Sample Input
    3 3 2 1 7 3 3 2 1 5 3 1 4 1 5
     
    Sample Output
    YES YES NO

    题意:给你一组数,能否删除一个后使他成为升序或者降序

    思路:

    正反分别求一次最长上升子序列,只要长度有一次≥n-1即可


    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <queue>
    #include <vector>
    #include <algorithm>
    #include <functional>
    typedef long long ll;
    using namespace std;
    const int maxn = 1e5 + 5;
    int b[maxn];
    int n;
    int Search(int num,int low,int high)
    {
        int mid;
        while(low <= high)
        {
            mid = (low+high)/2;
            if(num >= b[mid])
                low= mid+1;
            else
                high = mid-1;
        }
        return low;
    }
    
    int fin(int *a)
    {
        int len,pos;
        b[1] = a[1];
        len = 1;
        for(int i = 2;i <= n;i++)
        {
            if(a[i] >= b[len])
            {
                len = len+1;
                b[len] = a[i];
            }
            else
            {
                pos = Search(a[i],1,len);
                b[pos ]  = a[i];
            }
        }
        if(len >= n-1)
            return true;
        else
            return false;
    }
    
    int t[maxn],tt[maxn];
    
    
    int main()
    {
        int cas;
        scanf("%d",&cas);
        while(cas--)
        {
            scanf("%d",&n);
            for(int i = 1;i <= n;i++)
                scanf("%d",t+i);
            for(int i = 1;i <= n;i++)
                tt[n+1-i] = t[i];
            bool flag = fin(t);
            flag |= fin(tt);
            if(flag)
                printf("YES
    ");
            else
                printf("NO
    ");
        }
    }
    

      

  • 相关阅读:
    单例模式
    抽象工厂模式
    工厂方法模式
    并发编程:等待/通知机制
    并发编程:深入理解synchronized
    并发编程:为何线程的等待方法都会抛出InterruptException
    并发编程:基础知识
    Mybatis:整合Spring
    Mybatis:插件原理
    Mybatis:体系结构和工作原理
  • 原文地址:https://www.cnblogs.com/Przz/p/5409688.html
Copyright © 2020-2023  润新知