1004 array array array

array array array

Problem Description

One day, Kaitou Kiddo had stolen a priceless diamond ring. But detective Conan blocked Kiddo's path to escape from the museum. But Kiddo didn't want to give it back. So, Kiddo asked Conan a question. If Conan could give a right answer, Kiddo would return the ring to the museum.
Kiddo: "I have an array A and a number k, if you can choose exactly k elements from A and erase them, then the remaining array is in non-increasing order or non-decreasing order, we sayA is a magic array. Now I want you to tell me whether A is a magic array. " Conan: "emmmmm..." Now, Conan seems to be in trouble, can you help him?

The first line contains an integer T indicating the total number of test cases. Each test case starts with two integersn and k in one line, then one line with n integers: A1,A2…An.
1≤T≤20
1≤n≤105
0≤k≤n
1≤Ai≤105

For each test case, please output "A is a magic array." if it is a magic array. Otherwise, output "A is not a magic array." (without quotes).

34 11 4 3 75 24 1 3 1 26 11 4 3 5 4 6

A is a magic array.
A is a magic array.
A is not a magic array.

       思路： 一个水题然而在一个小时的时候才写完，确实不应该，本来一开始就有思路的然而不是敢写，最后还无脑开小了数组  T了两发。

       其实就是找一个最长非递增子序列和最长非递减子序列。用  o(  nlogn ) 的写法。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<string>#include<cstdio>#include<cstring>#include<cmath>#include<iostream>#include<queue>#include<stack>#include<vector>#include<set>#include<algorithm>#define maxn 10010#define INF 0x3f3f3f3f#define eps 1e-8#define MOD 1000000007#define ll long longusing namespace std;int a[100050];int d1[100050],d2[100050];int b[100050];int main(){    int T,n,k;    scanf("%d",&T);    while(T--)    {        scanf("%d %d",&n,&k);        for (int i=1;i<=n;i++) scanf("%d",&a[i]), b[n-i+1]=a[i];        d1[1]=a[1];        d2[1]=a[1];        int len1=1,len2=1;        for (int i=2;i<=n;i++)        {            if (a[i]>=d1[len1]) d1[++len1]=a[i];            else  //否则就找一个最该替换的替换掉            {                int j=upper_bound(d1+1,d1+len1+1,a[i])-d1;  //找到第一个大于它的d的下标                d1[j]=a[i];            }            if (b[i]>=d2[len2]) d2[++len2]=b[i];  //如果可以接在len后面就接上            else  //否则就找一个最该替换的替换掉            {                int j=upper_bound(d2+1,d2+len2+1,b[i])-d2;  //找到第一个大于它的d的下标                d2[j]=b[i];            }        }        int mm=min(n-len1,n-len2);        if(mm<=k)            puts("A is a magic array.");        else            puts("A is not a magic array.");    }        return 0;}