Educational Codeforces Round 99 (Rated for Div. 2) D. Sequence and Swaps（貪心）

題目連結：https://codeforc.es/contest/1455/problem/D

You are given a sequence a consisting of n integers a1,a2,…,an, and an integer x. Your task is to make the sequence a sorted (it is considered sorted if the condition a1≤a2≤a3≤⋯≤an holds).

To make the sequence sorted, you may perform the following operation any number of times you want (possibly zero): choose an integer i such that 1≤i≤n and ai>x, and swap the values of ai and x.

For example, if a=[0,2,3,5,4], x=1, the following sequence of operations is possible:

choose i=2 (it is possible since a2>x), then a=[0,1,3,5,4], x=2;
choose i=3 (it is possible since a3>x), then a=[0,1,2,5,4], x=3;
choose i=4 (it is possible since a4>x), then a=[0,1,2,3,4], x=5.
Calculate the minimum number of operations you have to perform so that a becomes sorted, or report that it is impossible.

Input
The first line contains one integer t (1≤t≤500) — the number of test cases.

Each test case consists of two lines. The first line contains two integers n and x (1≤n≤500, 0≤x≤500) — the number of elements in the sequence and the initial value of x.

The second line contains n integers a1, a2, …, an (0≤ai≤500).

The sum of values of n over all test cases in the input does not exceed 500.

Output
For each test case, print one integer — the minimum number of operations you have to perform to make a sorted, or −1, if it is impossible.

Example

input

6
4 1
2 3 5 4
5 6
1 1 3 4 4
1 10
2
2 10
11 9
2 10
12 11
5 18
81 324 218 413 324

output

3
0
0
-1
1
3

---

題意

對於一個長度為 n 的陣列，每次步驟可以將一個元素 a[i] 與 x 互換（前提是 ai > x），求至少需要換幾次使得數列成為非遞減序列。

分析

我們發現每次交換以後都只能將一個位置上的數變小，並且一旦交換，這個位置的數就不能再變了（因為不比 x 大了），所以可以推斷出，如果連續下降序列大於 2 ，就不可能通過操作使他變成非遞減的了。
對於一個位置 i ，若 a[i] < a[i+1] ，那這個點就必須要通過交換來變小，如果交換會使得 a[i - 1] > a[i] 了，那麼要先交換前面的一系列點，使得交換過後仍是非遞減數列。

程式碼

#include<bits/stdc++.h>
using namespace std;

int t;
int n,x;
int a[507];
int f[507];

int main()
{
	scanf("%d",&t);
	while(t--)
	{
		int ans = 0;
		scanf("%d%d",&n,&x);
		for(int i=1;i<=n;i++) scanf("%d",&a[i]), f[i] = 0;
		int flag = 1;
		for(int i=1;i<n;i++)
		{
			if(a[i] > a[i + 1])
			{
				if(f[i - 1] == 1) flag = 0;
				f[i] = 1;
			}
		}
		if(flag == 0)
		{
			printf("-1\n");
			continue;
		}
		for(int i=1;i<n;i++)
		{
			if(f[i] == 0) continue;
			int j = i;
			while(a[j - 1] > x && j - 1 > 0) j--;
			if(j == i && a[i] <= x)
			{
				flag = 0;
				break;
			}
			for(int k=j;k<=i;k++)
			{
				if(x == a[k]) continue;//注意這邊如果相等的不用交換 
				swap(x, a[k]);
				ans++;
			}
			if(a[i] > a[i + 1])
			{
				flag = 0;
				break;
			}
		}
		if(flag == 1) printf("%d\n",ans);
		else printf("-1\n");
	}
	return 0;
}