Uva11175 digraph D and E from D to e and back

subject

link

https://www.luogu.com.cn/problem/UVA11175

Literal description

D It's a digraph , D Each side of corresponds to E A node of , if D Have edge uv, be E Node uv, about D The two sides of uv and vw, E Two nodes in uv And vw Even the edge . give E, Judge whether there is a corresponding D.

Ideas

I didn't understand the question for a long time at the beginning , Refer to the solution , Briefly talk about ideas

1. enumeration 3 A little bit
2. If the first two points have edges to the third point , Record 1
3. If only one has , Record 2
4. At the end of the third cycle, judge the record 1 And record 2 Does it exist at the same time
   At the same time , error
   otherwise , continue
   This is because , Two points can go to multiple points at the same time , But suddenly there is a point that only one point can reach , This is the picture, so it can't be restored , So judge .
5. Finally, there was no error , correct

Code implementation

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

const int maxn=300+10;
int n,m,k;
bool vis[maxn][maxn];
inline bool solve(){
    
	for(int i=0;i<m;i++){
    
		for(int j=0;j<m;j++){
    
			bool flag1=false,flag2=false;
			for(int k=0;k<m;k++){
    
				if(vis[i][k]&&vis[j][k])flag1=true;
				if(vis[i][k]^vis[j][k])flag2=true;
			}
			if(flag1&&flag2)return false;
		}
	}
	return true;
}
int main(){
    
	scanf("%d",&n);
	int N=n;
	while(n--){
    
		printf("Case #%d: ",N-n);
		memset(vis,false,sizeof(vis)); 
		scanf("%d%d",&m,&k);
		for(int i=1;i<=k;i++){
    
			int x,y;
			scanf("%d%d",&x,&y);
			vis[x][y]=true;
		}
		if(solve())printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}