求组合实际上可以不用标记走过，可以利用下标来，往后走就是

#include<cstdio>
#include<iostream>
#include<algorithm> 
using namespace std;
int num[12];
int mstate[12];
int n,r;
void dfs(int top,int numi){
	if(top==r){
		for(int k=0;k<top-1;++k) cout<<mstate[k]<<" ";
		cout<<mstate[top-1]<<endl;
		return;
	}

	for(int i=numi;i<n;i++){
		mstate[top]=num[i];
		dfs(top+1,i+1);
	}
	return;
}
int main(){

	cin>>n>>r;
	for(int i=0;i<n;i++) cin>>num[i];
	sort(num,num+n);
	dfs(0,0);	
	return 0;
	
}

顺便写了个组合并且全排列，这时候就需要flag记录了

#include<cstdio>
#include<iostream>
#include<algorithm> 
using namespace std;
int num[12];
int flag[12];
int mstate[12];
int n,r;
int mcount=0;
void dfs(int top){
	if(top==r){
		for(int k=0;k<top-1;++k) cout<<mstate[k]<<" ";
		cout<<mstate[top-1]<<endl;
		mcount++;
		return;
	}

	for(int i=0;i<n;i++){
		if(flag[i]!=1){
		
		mstate[top]=num[i];
		flag[i]=1;
		dfs(top+1);
		flag[i]=0;
		}
	}
	return;
}
int main(){

	cin>>n>>r;
	for(int i=0;i<n;i++) cin>>num[i];
	sort(num,num+n);
	dfs(0);
	cout<<mcount;	
	return 0;
	
}

#include<bits/stdc++.h>
#define int long long 
#define endl '\n'
using namespace std;
int n,r;
int nums[10];
bool st[10];
int ans[10];
bool flag = true;
void dfs(int accounts){
	if(accounts == r){
		for(int i = 0; i < r - 1;i++){
			if(ans[i] > ans[i+1]){
			flag = false;
		}
		}
		if(flag == true){
		for(int i = 0;i < r;i++){
			cout<<ans[i]<<" ";
		}
		cout<<endl;
	}
	flag = true;
		return ;
	}
	for(int i = 0;i < n;i++){
		if(st[i] == false){
		ans[accounts] = nums[i];
		st[i] = true;
		dfs(accounts+1);
		st[i] = false;
		}
	}
	return ;
}
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin>>n;
	cin>>r;
	for(int i = 0;i < n;i++){
		cin>>nums[i];
	}
	sort(nums,nums+n);
	dfs(0);
	return 0;
}

import java.util.*;

public class Main {
	public static boolean flag[];// 标记第i个元素是否被使用过
	public static int n;
	public static int road[];

	//用来检查是否是正序的
	public static boolean check(int road[], int total) {
		boolean natural = true;
		for (int i = 1; i < total; i++) {
			if (road[i] > road[i + 1])
				natural = false;
		}
		return natural;
	}

	// idx 选择第几个数了
	// 选取元素总数
	public static void dfs(int idx, int num[], int total) {
		if (idx > total) {
			if (check(road,total)) {
				for (int i = 1; i <= total; i++) {
					System.out.print(road[i] + " ");
				}
				System.out.println();
			}
			return;
		}
		for (int i = idx ; i <= n; i++) {
			if (!flag[i]) {
				road[idx] = num[i];
				flag[i] = true;
				dfs(idx + 1, num, total);
				flag[i] = false;
			}
		}
		return;
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		int r = sc.nextInt();

		flag = new boolean[n + 5];
		int num[] = new int[n + 5];
		road = new int[n + 5];
		for (int i = 1; i <= n; i++) {
			num[i] = sc.nextInt();
		}
//		fromIndex 是排序开始的位置（包含），toIndex 是排序结束的位置（不包含）。
//		这意味着 Arrays.sort(a, fromIndex, toIndex) 将对数组 a 从索引 fromIndex 到 toIndex - 1 的范围进行排序。
		Arrays.sort(num, 1, n+1);
		sc.close();
		dfs(1, num, r);
	}
}

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<climits>
#include<vector>
using namespace std;
int n,r;
vector<int> res;
void dfs(int startIndex,int countNum,int nums[],int used[]){
//	处理递归终止的条件
	if(countNum>r){
		for(auto temp:res) cout<<temp<<" ";
		cout<<endl;
		return;
	}
//	处理递归的单层逻辑
	for(int i=startIndex;i<n;i++){
		if(used[i]==1) continue;
		used[i]=1;
		res.push_back(nums[i]);
		dfs(i+1,countNum+1,nums,used);
		used[i]=0;	
		res.pop_back();
		
	}
}
int main(){
	cin>>n>>r;
	int nums[n];
	int used[n]={0};
	for(int i=0;i<n;i++) scanf("%d",&nums[i]);
	sort(nums,nums+n);
	dfs(0,1,nums,used);//从第一个数开始选
	return 0;
}

DFS即可

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