QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#232610#7640. Colorful Cycles飞带长队 (Xintong Fang, Jun Zheng, Haifeng Liu)TL 12ms125824kbC++142.2kb2023-10-30 17:31:422023-10-30 17:31:42

Judging History

This is the latest submission verdict.

  • [2024-07-04 22:58:32]
  • hack成功,自动添加数据
  • (/hack/728)
  • [2023-10-30 17:31:42]
  • Judged
  • Verdict: TL
  • Time: 12ms
  • Memory: 125824kb
  • [2023-10-30 17:31:42]
  • Submitted

answer

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
#ifdef DEBUG
template<typename T>
ostream& operator << (ostream &out,const vector<T> &x){
	if(x.empty())return out<<"[]";
	out<<'['<<x[0];
	for(int i=1,len=x.size();i<len;i++)out<<','<<x[i];
	return out<<']';
}
template<typename T>
vector<T> ary(const T *a,int l,int r){
	return vector<T>{a+l,a+1+r};
}
template<typename T>
void debug(T x){
	cerr<<x<<endl;
}
template<typename T,typename ... S>
void debug(T x,S...y){
	cerr<<x<<' ',debug(y...);
}
#endif
const int N=1e6+10,V=N*2;
int T,n,m,k;
vector<pair<int,int> >A[N];
vector<int>B[V];
vector<tuple<int,int,int> >E[V];
int dft,top,dfn[N],low[N],stk[N];
int fa[V];
void add(int u,int v){
	// debug(u,v);
	B[u].push_back(v),B[v].push_back(u);
}
#define v e.first
#define w e.second
void tarjan(int u,int fa=0){
	dfn[u]=low[u]=++dft,stk[++top]=u;
	for(auto e:A[u])if(v^fa){
		if(!dfn[v]){
			tarjan(v),low[u]=min(low[u],low[v]);
			if(low[v]>=dfn[u]){
				add(++k,u);
				do add(k,stk[top]);while(stk[top--]^v);
			}
		}else low[u]=min(low[u],dfn[v]);
	}
}
void build(){
	for(int u=1;u<=n;u++){
		for(auto e:A[u]){
			if(u>v)continue;
			if(fa[u]==fa[v]||fa[fa[u]]==v)E[fa[u]].push_back({u,v,w});
			else E[fa[v]].push_back({u,v,w});
		}
	}
}
#undef v
#undef w
void dfs(int u){
	for(int v:B[u])if(v^fa[u]){
		fa[v]=u,dfs(v);
	}
}
int col[N];
bool get(){
	scanf("%d%d",&n,&m),k=n;
	for(int u,v,w;m--;){
		scanf("%d%d%d",&u,&v,&w);
		A[u].push_back({v,w}),A[v].push_back({u,w});
	}
	for(int i=1;i<=n;i++)if(!dfn[i])tarjan(i);
	for(int i=1;i<=k;i++)if(!fa[i])dfs(i);
	build();
	for(int i=n+1;i<=k;i++){
		int sum=0;
		for(int v:B[i])col[v]=0;
		for(auto e:E[i]){
			int u,v,w;
			tie(u,v,w)=e;
			col[u]|=1<<w,col[v]|=1<<w,sum|=1<<w;
		}
		if(__builtin_popcount(sum)!=3)continue;
		int cnt=0;
		for(int v:B[i]){
			cnt+=__builtin_popcount(col[v])>1;
		}
		if(cnt>2)return 1;
	}
	return 0;
}
void clr(){
	dft=top=0;
	for(int i=1;i<=n;i++){
		A[i].clear();
		dfn[i]=low[i]=0;
	}
	for(int i=1;i<=k;i++){
		B[i].clear(),E[i].clear();
		fa[i]=0;
	}
}
int main(){
	for(scanf("%d",&T);T--;)puts(get()?"Yes":"No");
	return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 12ms
memory: 125824kb

input:

2
3 3
1 2 3
2 3 1
1 3 2
5 6
1 2 1
2 3 1
1 3 2
3 4 3
3 5 3
4 5 3

output:

Yes
No

result:

ok 2 token(s): yes count is 1, no count is 1

Test #2:

score: -100
Time Limit Exceeded

input:

100000
7 10
7 2 2
6 4 2
6 1 2
7 1 3
3 4 1
6 7 1
2 6 3
3 1 2
5 3 1
2 1 1
7 10
5 7 3
7 1 1
4 6 3
6 3 1
3 4 3
4 2 2
3 2 3
1 3 3
3 7 1
1 4 2
7 10
5 6 3
3 5 2
7 2 3
7 3 3
1 2 2
4 3 2
7 4 2
6 1 2
2 6 1
7 5 2
7 10
7 1 3
7 5 3
6 4 1
7 6 1
1 4 1
3 4 2
2 7 2
1 3 1
3 5 3
5 1 3
7 10
6 7 2
3 4 3
1 4 2
5 3 2
7 4 ...

output:

Yes
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No...

result: