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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#232610 | #7640. Colorful Cycles | ucup-team1004 | TL | 12ms | 125824kb | C++14 | 2.2kb | 2023-10-30 17:31:42 | 2023-10-30 17:31:42 |
Judging History
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;
}
详细
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...