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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#860873 | #9678. 网友小 Z 的树 | XY_Eleven | 0 | 3ms | 10180kb | C++23 | 4.8kb | 2025-01-18 15:22:06 | 2025-01-18 15:24:33 |
Judging History
answer
#include <bits/stdc++.h>
#include "diameter.h"
// #include <windows.h>
// #include <bits/extc++.h>
// using namespace __gnu_pbds;
using namespace std;
//#pragma GCC optimize(3)
#define DB double
#define LL long long
#define ULL unsigned long long
#define in128 __int128
#define cint const int
#define cLL const LL
#define For(z,e1,e2) for(int z=(e1);z<=(e2);z++)
#define Rof(z,e1,e2) for(int z=(e2);z>=(e1);z--)
#define For_(z,e1,e2) for(int z=(e1);z<(e2);z++)
#define Rof_(z,e1,e2) for(int z=(e2);z>(e1);z--)
#define inint(e) scanf("%d",&e)
#define inll(e) scanf("%lld",&e)
#define inpr(e1,e2) scanf("%d%d",&e1,&e2)
#define in3(e1,e2,e3) scanf("%d%d%d",&e1,&e2,&e3)
#define outint(e) printf("%d\n",e)
#define outint_(e) printf("%d%c",e," \n"[i==n])
#define outint2_(e,e1,e2) printf("%d%c",e," \n"[(e1)==(e2)])
#define outll(e) printf("%lld\n",e)
#define outll_(e) printf("%lld%c",e," \n"[i==n])
#define outll2_(e,e1,e2) printf("%lld%c",e," \n"[(e1)==(e2)])
#define exc(e) if(e) continue
#define stop(e) if(e) break
#define ret(e) if(e) return
#define ll(e) (1ll*(e))
#define pb push_back
#define ft first
#define sc second
#define pii pair<int,int>
#define pli pair<long long,int>
#define vct vector
#define clean(e) while(!e.empty()) e.pop()
#define all(ev) ev.begin(),ev.end()
#define sz(ev) ((int)ev.size())
#define debug(x) printf("%s=%d\n",#x,x)
#define x0 __xx00__
#define y1 __yy11__
#define ffo fflush(stdout)
cLL mod=998244353,G=404;
// cLL mod[2]={1686688681ll,1666888681ll},base[2]={166686661ll,188868881ll};
template <typename Type> void get_min(Type &w1,const Type w2) { if(w2<w1) w1=w2; } template <typename Type> void get_max(Type &w1,const Type w2) { if(w2>w1) w1=w2; }
template <typename Type> Type up_div(Type w1,Type w2) { return (w1/w2+(w1%w2?1:0)); }
template <typename Type> Type gcd(Type X_,Type Y_) { Type R_=X_%Y_; while(R_) { X_=Y_; Y_=R_; R_=X_%Y_; } return Y_; } template <typename Type> Type lcm(Type X_,Type Y_) { return (X_/gcd(X_,Y_)*Y_); }
template <typename Type> Type md(Type w1,const Type w2=mod) { w1%=w2; if(w1<0) w1+=w2; return w1; } template <typename Type> Type md_(Type w1,const Type w2=mod) { w1%=w2; if(w1<=0) w1+=w2; return w1; }
void ex_gcd(LL &X_,LL &Y_,LL A_,LL B_) { if(!B_) { X_=1ll; Y_=0ll; return ; } ex_gcd(Y_,X_,B_,A_%B_); X_=md(X_,B_); Y_=(1ll-X_*A_)/B_; } LL inv(LL A_,LL B_=mod) { LL X_=0ll,Y_=0ll; ex_gcd(X_,Y_,A_,B_); return X_; }
template <typename Type> void add(Type &w1,const Type w2,const Type M_=mod) { w1=md(w1+w2,M_); } void mul(LL &w1,cLL w2,cLL M_=mod) { w1=md(w1*md(w2,M_),M_); } template <typename Type> Type pw(Type X_,Type Y_,Type M_=mod) { Type S_=1; while(Y_) { if(Y_&1) mul(S_,X_,M_); Y_>>=1; mul(X_,X_,M_); } return S_; }
template <typename Type> Type bk(vector <Type> &V_) { auto T_=V_.back(); V_.pop_back(); return T_; } template <typename Type> Type tp(stack <Type> &V_) { auto T_=V_.top(); V_.pop(); return T_; } template <typename Type> Type frt(queue <Type> &V_) { auto T_=V_.front(); V_.pop(); return T_; }
template <typename Type> Type bg(set <Type> &V_) { auto T_=*V_.begin(); V_.erase(V_.begin()); return T_; } template <typename Type> Type bk(set <Type> &V_) { auto T_=*prev(V_.end()); V_.erase(*prev(V_.end())); return T_; }
mt19937 gen(time(NULL)); int rd() { return abs((int)gen()); }
int rnd(int l,int r) { return rd()%(r-l+1)+l; }
void main_init()
{
}
map <array<int,3>,int> mp;
int qry(vct <int> w)
{
sort(all(w));
if(mp.count({w[0],w[1],w[2]})) return mp[{w[0],w[1],w[2]}];
return (mp[{w[0],w[1],w[2]}]=query(w[0],w[1],w[2]));
}
void operator += (array <int,3> &w1,array <int,3> w2)
{
vct <int> h={w1[0],w1[1],w1[2],w2[0],w2[1],w2[2]};
sort(all(h)); h.erase(unique(all(h)));
int len=sz(h),mx=0;
For_(i,0,len) For_(j,i+1,len) For_(k,j+1,len)
{
int t=qry({h[i],h[j],h[k]});
if(t>=mx)
w1={h[i],h[j],h[k]},mx=t;
}
}
std::pair<int, int> find_diameter(int task_id, int n)
{
if(n==1) return {1,1};
if(n==2) return {1,2};
mp.clear();
vct <int> a(n);
For_(i,0,n) a[i]=i+1;
shuffle(all(a),gen);
vct <array<int,3> > g;
for(int i=0;i<n;i+=3)
{
if(i+2<n) g.pb({a[i],a[i+1],a[i+2]});
else g.pb({a[n-1],a[n-2],a[n-3]});
}
int m=sz(g);
For_(i,1,m) g[0]+=g[i];
auto [w1,w2,w3]=g[0];
auto dis=[&](int x,int y)->array<int,2>
{
array <int,2> mn={n<<2|11,0};
For(i,1,n) if(i!=x&&i!=y)
get_min(mn,{qry({x,y,i}),i});
return mn;
};
auto t1=dis(w1,w2),t2=dis(w2,w3),t3=dis(w1,w3),t=max({t1,t2,t3});
if(t2==t) w1=w3;
else if(t3==t) w2=w3;
w3=t[1];
// printf("%d,%d,%d\n",w1,w2,w3);
if(in(w1,w2,w3)) return {w2,w3};
if(in(w2,w1,w3)) return {w1,w3};
return {w1,w2};
}
Details
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Subtask #1:
score: 0
Wrong Answer
Test #1:
score: 0
Wrong Answer
time: 3ms
memory: 10180kb
input:
1 100 25 1 3 2 18 3 8 4 18 5 14 6 22 7 18 8 10 9 11 10 12 11 25 12 16 13 11 14 17 15 17 16 25 17 2 18 20 19 18 20 12 21 1 22 17 23 14 24 1 50 1 37 2 27 3 10 4 25 5 16 6 17 7 10 8 36 9 16 10 6 11 48 12 2 13 28 14 30 15 10 16 44 17 31 18 1 19 6 20 7 21 30 22 42 23 45 24 23 25 27 26 39 27 45 28 48 29 4...
output:
WA
result:
wrong answer Wrong Answer
Subtask #2:
score: 0
Skipped
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