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.
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
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
#include
#include
#include
#include
#include
#include
#include
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\n");
else
printf("NO\n");
}
}