HDU 5532 2015ACM-ICPC长春赛区现场赛F题

Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 3582 Accepted Submission(s): 908

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 a_1, a_2, \ldots, a_n, 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 a_1, a_2, \ldots, a_n.

1 \le T \le 2000

2 \le n \le 10^5

1 \le a_i \le 10^5

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

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

题意：去掉一个数，问是否能使序列变成单调的

思路：之前一直是用模拟做的，其实可以直接算一下LIS，正着反着都算一遍，O（ologn）

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int main() {
int T, n, i, j, k, a[100005], k1, k2, ans[100005], len, b[100005], flag;
scanf("%d", &T);
while (T--) {
flag = 0;
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
memset(ans, 0, sizeof(ans));
k1 = k2 = 0;
scanf("%d", &n);
for (i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
ans[1] = a[1];
len = 1;
for (i = 2; i <= n; i++) {
if (a[i] >= ans[len]) {
ans[++len] = a[i];
} else {
int pos = upper_bound(ans + 1, ans + len + 1, a[i]) - ans;
ans[pos] = a[i];
}
}
if (len == n - 1 || len == n) {
printf("YES\n");
continue;
}
for (i = 1; i <= n; i++) {
b[i] = a[n - i + 1];
}
ans[1] = b[1];
len = 1;
for (i = 2; i <= n; i++) {
if (b[i] >= ans[len]) {
ans[++len] = b[i];
} else {
int pos = upper_bound(ans + 1, ans + len + 1, b[i]) - ans;
ans[pos] = b[i];
}
}
if (len == n - 1 || len == n)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}