JakeLin- [Blue Bridge Cup] [Algorithm Training VIP] Competition Arrangement-Problem Solving

Title description

There are 2 ^ n (n <= 6) teams playing single-cycle games, which are planned to be completed within 2 ^ n–1 days, and each team plays one game every day. Design a game schedule so that each team competes with different opponents within 2 ^ n–1 days. 

Input

A line, enter the value of n. 

Output

There are a total of (2 ^ n – 1) lines. The ith line outputs the schedule of the ith day. 
The format is: <i> AB CD where i is the number of days, A and B are the numbers of the two sides of the game, each row has a total of 2n-1 matches. 

Sample input

2 

Sample output

<1>1-2 3-4
<2>1-3 2-4
<3>1-4 2-3 

Original title link: [Blue Bridge Cup] [Algorithm Training VIP] Competition Arrangement

Idea:
Take n = 3 as an example, a total of 8 people, 7 days, then the first column must be 1 and 7 other people.
The next two have not played in groups of two, but note that if the two of them have played a few days ago, do not arrange for them to play together, so use the viss array to record who played with whom .
[Adopt the idea of ​​linked list traversal, but judge directly here]

Description of two vis array variables:

name	effect
vis[100]	Throughout the line , vis [j] indicates that j has been hit this day, please do not arrange him
all [100]	Throughout the entire program , viss [i] [j] indicates that "ij" has been hit, please do not arrange them together

 Attached code:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
using namespace std;
int vis[100];
int viss[100][100];
int main(){
    int n;
    cin>>n;
    int per = pow(2,n);
    int day = pow(2,n)-1;
    for(int i=1;i<=day;i++){
        memset(vis,0,sizeof(vis));    
        cout<<"<"<<i<<">";
        int p=1;
        while(p<=per){  
            if(vis[p]==1){  //找出第一个还没打过的p
                p++;
                continue;
            }
            int tmp=p;
            for(int q=p+1;q<=per;q++){ //判断谁和p打：找出q
                if(vis[q]==0 && viss[p][q]==0){  //如果他们之前没打过
                    vis[p]=vis[q]=viss[p][q]=1;
                    cout<<p<<"-"<<q<<" ";
                    p=tmp+1;   //判断下一个人p
                    break;
                }
            }
        }
        cout<<endl;
    }
    return 0;
}