leetcode-634 Find the Derangement of An Array（水水的模拟）

Given an array containing a permutation of 1 to n, you have to find the minimum number of swaps to sort the array in ascending order. A swap means, you can exchange any two elements of the array.

For example, let n = 4, and the array be 4 2 3 1, then you can sort it in ascending order in just 1 swaps (by swapping 4 and 1).

Input
Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains two lines, the first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers separated by spaces. You may assume that the array will always contain a permutation of 1 to n.

Output
For each case, print the case number and the minimum number of swaps required to sort the array in ascending order.

Sample Input
3

4

4 2 3 1

4

4 3 2 1

4

1 2 3 4

Sample Output
Case 1: 1

Case 2: 2

Case 3: 0

题解：

直接模拟全过程就好了

代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<stdio.h>
#include<math.h>
#include<string>
#include<stdio.h>
#include<queue>
#include<stack>
#include<map>
#include<deque>
#define M (t[k].l+t[k].r)/2
#define lson k*2
#define rson k*2+1
using namespace std;
int a[105];
int b[105];
int main()
{
int i,j,k,ans,num=1,n,t,c;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
c=0;
ans=0;
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
b[i]=a[i];
}
sort(b,b+n);
for(i=0;i<n;i++)//对照排序以后的来模拟交换
{
if(a[i]==b[i])
continue;
for(j=i+1;j<n;j++)
{
if(a[j
4000
]==b[i])
{
swap(a[i],a[j]);
c++;
break;
}
}
}
printf("Case %d: %d\n",num,c);
num++;
}
return 0;
}