zoukankan      html  css  js  c++  java
  • HDU 5532 Almost Sorted Array

    Almost Sorted Array

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
    Total Submission(s): 831    Accepted Submission(s): 293


    Problem Description

    We are all familiar with sorting algorithms: quick sort, merge sort, heap sort, insertion sort, selection sort, bubble sort, etc. But sometimes it is an overkill to use these algorithms for an almost sorted array.

    We say an array is sorted if its elements are in non-decreasing order or non-increasing order. We say an array is almost sorted if we can remove exactly one element from it, and the remaining array is sorted. Now you are given an array a1,a2,…,an, is it almost sorted?

    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.

    1≤T≤2000
    2≤n≤105
    1≤ai≤105
    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
     
    Source
     
    解题:直接拿LIS刷几遍就是了
     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 const int maxn = 100010;
     4 const int INF = 0x3f3f3f3f;
     5 int a[maxn],d[maxn];
     6 int main() {
     7     int kase,n;
     8     scanf("%d",&kase);
     9     while(kase--) {
    10         scanf("%d",&n);
    11         memset(d,0x3f,sizeof d);
    12         for(int i = 0; i < n; ++i) {
    13             scanf("%d",a + i);
    14             *upper_bound(d,d + n,a[i]) = a[i];
    15         }
    16         auto ret = lower_bound(d,d + n,INF) - d;
    17         memset(d,0x3f,sizeof d);
    18         for(int i = 0; i < n; ++i)
    19             *upper_bound(d,d + n,-a[i]) = -a[i];
    20         ret = max(ret,lower_bound(d,d + n,INF) - d);
    21         printf("%s
    ",ret >= n - 1?"YES":"NO");
    22     }
    23     return 0;
    24 }
    View Code
  • 相关阅读:
    qt 问题及处理
    windows 依赖查看
    java基础知识
    Linux 相关
    OutLook中添加Exchange失败问题
    开源协议简介
    grail开发环境的搭建
    node+mongodb+WP构建的移动社交应用源码 分享
    INotifyPropertyChanged接口的详细说明
    WP8.1开发:后台任务详解(求推荐)
  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4957216.html
Copyright © 2011-2022 走看看