using namespace std;
int main(){
	int n;
	cin>>n;
	int a[1010];
	for(int i=1; i<=n; i++){
		a[i]=i;
	}
	do{
		for(int i=1; i<n; i++){
			printf("%d ",a[i]);
		}
		printf("%d\n",a[n]);
	}while(next_permutation(a+1,a+n+1));//直接用内置的全排列函数写就行了，注意先输出已有的第一中排列，所以要用do-while
	return 0;
}

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

int num[10],str[10]; int ans=0; void dfs(int n,int n2){

for(int i=1;i<=n2;i++) { if(str[i]!=0) continue; str[i] =1; num[n]=i; dfs(n+1,n2); str[i]=0; }

if(n==n2) { for(int i=1;i<=n2;i++) { cout<<num[i]<<' '; } cout<<endl; }

} int main() { int n2; cin>>n2; dfs(1,n2); return 0; }

#include <bits/stdc++.h>
using namespace std;
int n = 0;
vector<int> v;

int main(){
	cin >> n;
	for(int i = 1 ; i <= n ; i++){
		v.push_back(i);
	}
	
	do{
		for(int i = 0 ; i <n ; i++){
			cout << v[i] << " ";
		}
		cout << endl;	
	}	while(next_permutation(v.begin(), v.end()));
		
	return 0;
}

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<vector>
using namespace std;
const int N=12;
int used[N];
int nums[N];
vector<int> res;
int n;
void dfs(int countNum){
	if(countNum>n){
		for(auto temp:res) cout<<temp<<" ";
		cout<<endl;
		return;
	}
	for(int i=0;i<n;i++){
		if(used[i]) continue;
//		标记为被使用过
		used[i]=1;
		res.push_back(nums[i]);
		dfs(countNum+1);
		used[i]=0;
		res.pop_back();
	}
}
int main(){

	cin>>n;
	for(int i=0;i<n;i++) nums[i]=i+1;
	dfs(1);
	
	return 0;
}

提供两种解法：

next_permutation

#include <bits/stdc++.h>
using namespace std;
const int N = 20;
int a[N];
int n;
int main(){
    scanf("%d",&n);
    for (int i = 1; i <= n; ++i) {
        a[i] = i;
    }
    do{
        for (int i = 1; i <= n; ++i) {
            printf("%d ",a[i]);
        }
        printf("\n");
    }while(next_permutation(a + 1,a + n + 1));
    return 0;
}

DFS

#include <bits/stdc++.h>
using namespace std;
const int N = 20;
int vis[N];
int a[N];
int n;
void dfs(int x){
    if (x > n) {
        for (int i = 1; i <= n; ++i) {
            printf("%d ",a[i]);
        }
        printf("\n");
    }
    for (int i = 1; i <= n; ++i) {
        if(!vis[i]) {
            vis[i] = 1;
            a[x] = i;
            dfs(x + 1);
            a[x] = 0;
            vis[i] = 0;
        }
    }
}
int main(){
    scanf("%d",&n);
    dfs(1);
    return 0;
}

n = int(input())
nums = [i for i in range(1,n+1)]
res = []
path = []
used = [0]*n
def dfs(nums,path,used,n):
    if len(path) == n:
        print(*path)

    for i in range(n):
        if used[i]:
            continue
        used[i] = 1
        path.append(nums[i])
        dfs(nums,path,used,n)
        path.pop()
        used[i] = 0

dfs(nums,path,used,n)

#include <bits/stdc++.h> using namespace std; int a[20], x, n;

int b[20];

void dfs(int x) { if (x > n) { for (int i = 1; i < n; i++) cout << a[i] << " "; cout << a[n] << endl; return; } for (int i = 1; i <= n ; i++) {

	if (b[i] == 0) {
		b[i] = 1;
		a[x] = i;
		dfs(x + 1);
		b[i] = 0;
	}
}

}

int main() { cin >> n; dfs(1); return 0; }

#include <cstdio>
#include <vector>
#include <functional>

using namespace std;

int main() {
	int n;
	scanf("%d", &n);
	
	vector<char> ita(n + 1, 0);
	vector<int> res;
	function<void()> dfs = [&]() {
		if (res.size() >= n) {
			for (int& it : res)
				printf("%d ", it);
			printf("\n");
			return;
		}
		
		for (int i = 1; i <= n; ++i) {
			if (!ita[i]) { // 是否已经放过
				ita[i] = 1;
				res.push_back(i); // 放
				dfs();
				res.pop_back(); // 不放
				ita[i] = 0;
			}
		}
	};
	
	dfs();
	
	return 0;
}