CF1455 D. Sequence and Swaps（模擬）

連結 https://codeforces.com/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
給你一個序列和x，每次可以將序列中一個比x的數和x交換，問你最少操作次數，時序列升序排序，如果不可能輸出-1。

思路
模擬，我們可以用一個新陣列複製一下，將x插入，對他進行排序，這樣我們知道排序結果，然後我們反推次數，如果交換之後還不能升序，這種肯定是不可能的，注意的是，有些情況下我們不用嚴格按照排序後的序列排序，因為我們可能某次操作後已經滿足了。
例如
5 3
1 2 5 3 6

程式碼

#include <bits/stdc++.h>
typedef long long ll;
const ll mod = 1e9+7;
namespace fastIO {
    inline void input(int& res) {
        char c = getchar();res = 0;int f = 1;
        while (!isdigit(c)) { f ^= c == '-'; c = getchar(); }
        while (isdigit(c)) { res = (res << 3) + (res << 1) + (c ^ 48);c = getchar(); }
        res = f ? res : -res;
    }
    inline ll qpow(ll a, ll b) {
        ll ans = 1, base = a;
        while (b) {
            if (b & 1) ans = (ans * base % mod +mod )%mod;
            base = (base * base % mod + mod)%mod;
            b >>= 1;
        }
        return ans;
    }
}
using namespace fastIO;
using namespace std;
const int N = 1e3 + 5;
int Case,n,x; 
int a[N],aa[N];
bool check(int x){
	for(int i=x;i<=n;i++){
		if(a[i]<a[i-1]) return false;
	}
	return true;
}
int main(){
	Case=1;
	scanf("%d",&Case);
	while(Case--){
		bool flag = true;
		bool f = true;
		scanf("%d %d",&n,&x);
		for(int i=1;i<=n;i++){
			scanf("%d",&a[i]);
			if(a[i]<a[i-1]) f=false;
			aa[i]=a[i];
		}
		aa[n+1]=x;
		sort(aa+1,aa+n+1+1);
		int cnt = 0;
		for(int i=1;i<=n;i++){
			if(a[i]==aa[i]);
			else{
				if(a[i]>x){
					swap(a[i],x);
					cnt++;
				}
				else
					flag = false;
				if(a[i]<a[i-1]) flag = false;
			}
			if(check(i)) break;
		}
		if(f) printf("0\n");
		else
			printf("%d\n",flag?cnt:-1);
 	}
	return 0;
}