CF1662F.Antennas
普及/提高-
通过率:0%
AC君温馨提醒
该题目为【codeforces】题库的题目,您提交的代码将被提交至codeforces进行远程评测,并由ACGO抓取测评结果后进行展示。由于远程测评的测评机由其他平台提供,我们无法保证该服务的稳定性,若提交后无反应,请等待一段时间后再进行重试。
题目描述
There are n equidistant antennas on a line, numbered from 1 to n . Each antenna has a power rating, the power of the i -th antenna is pi .
The i -th and the j -th antenna can communicate directly if and only if their distance is at most the minimum of their powers, i.e., ∣i−j∣≤min(pi,pj) . Sending a message directly between two such antennas takes 1 second.
What is the minimum amount of time necessary to send a message from antenna a to antenna b , possibly using other antennas as relays?
输入格式
Each test contains multiple test cases. The first line contains an integer t ( 1≤t≤100000 ) — the number of test cases. The descriptions of the t test cases follow.
The first line of each test case contains three integers n , a , b ( 1≤a,b≤n≤200000 ) — the number of antennas, and the origin and target antenna.
The second line contains n integers p1,p2,…,pn ( 1≤pi≤n ) — the powers of the antennas.
The sum of the values of n over all test cases does not exceed 200000 .
输出格式
For each test case, print the number of seconds needed to trasmit a message from a to b . It can be shown that under the problem constraints, it is always possible to send such a message.
输入输出样例
输入#1
3 10 2 9 4 1 1 1 5 1 1 1 1 5 1 1 1 1 3 1 3 3 3 1
输出#1
4 0 2
说明/提示
In the first test case, we must send a message from antenna 2 to antenna 9 . A sequence of communications requiring 4 seconds, which is the minimum possible amount of time, is the following:
- In 1 second we send the message from antenna 2 to antenna 1 . This is possible since ∣2−1∣≤min(1,4)=min(p2,p1) .
- In 1 second we send the message from antenna 1 to antenna 5 . This is possible since ∣1−5∣≤min(4,5)=min(p1,p5) .
- In 1 second we send the message from antenna 5 to antenna 10 . This is possible since ∣5−10∣≤min(5,5)=min(p5,p10) .
- In 1 second we send the message from antenna 10 to antenna 9 . This is possible since ∣10−9∣≤min(5,1)=min(p10,p9) .