# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
972043 | de_sousa | Harbingers (CEOI09_harbingers) | C++17 | 91 ms | 24440 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
#define all(x) begin(x),end(x)
template<class T>
using iset=tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
void solve(){
int n;
cin>>n;
vector<vector<array<long long,2>>> g(n);
for(int i=1;i<n;i++){
long long x,y,z;
cin>>x>>y>>z;
x--;
y--;
g[x].push_back({y,z});
g[y].push_back({x,z});
}
vector<array<long long,2>> a(n);
for(int i=1;i<n;i++){
cin>>a[i][0]>>a[i][1];
}
auto eval=[&](array<long long,2>&x,long long y)->long long
{
return x[0]*y+x[1];
};
auto cross=[&](array<long long,2>&x,array<long long,2>&y,array<long long,2>&z)->long long
{
array<long long,2> A{y[0]-x[0],y[1]-x[1]};
array<long long,2> B{z[0]-y[0],z[1]-y[1]};
__int128 L=A[0];
L*=B[1];
__int128 R=A[1];
R*=B[0];
return L>=R;
};
vector<long long> depth(n);
vector<array<long long,2>> S(n,{0,0});
int Ssize=1;
vector<long long> sol(n);
auto dfs=[&](auto&&dfs,int x,int p)->void
{
int oldSsize=Ssize;
array<long long,2> newVal;
{
// int Siter=0;
// while(Siter+1<Ssize && eval(S[Siter],a[x][1])>eval(S[Siter+1],a[x][1]))Siter++;
// sol[x]=depth[x]*a[x][1] + eval(S[Siter],a[x][1]) + a[x][0];
int L=0;
int R=Ssize-1;
while(L<R){
int mid=L+(R-L)/2;
if(eval(S[mid],a[x][1])>=eval(S[mid+1],a[x][1]))L=mid+1;
else R=mid;
}
sol[x]=depth[x]*a[x][1] + eval(S[L],a[x][1]) + a[x][0];
newVal={-depth[x],sol[x]};
}
{
// while(Ssize>=2 && cross(S[Ssize-2],S[Ssize-1],newVal))Ssize--;
// ^- gives TLE but correct
// we're gonna ask the question "can we pop?" everywhere we can, i.e.: [2,Ssize] ([2,Ssize+1))
// Ssize will be the last that answers no
int L=2;
int R=Ssize+1;
Ssize=1;
while(L<R){
int mid=L+(R-L)/2;
// if we put the value at mid, can it be popped?
if(cross(S[mid-2],S[mid-1],newVal)){ // yes
// Ssize must be below mid
R=mid;
}
else{ // no
// Ssize can be mid or above, since L is growing, let's put it right now
Ssize=mid;
L=mid+1;
}
}
}
auto oldVal=S[Ssize];
S[Ssize]=newVal;
Ssize++;
for(auto[y,z]:g[x]){
if(y==p)continue;
depth[y]=depth[x]+z;
dfs(dfs,y,x);
}
Ssize--;
S[Ssize]=oldVal;
Ssize=oldSsize;
};
for(auto[x,y]:g[0]){
depth[x]=y;
dfs(dfs,x,0);
}
for(int i=1;i<n;i++){
cout<<sol[i]<<" \n"[i==n-1];
}
}
int main(){
ios::sync_with_stdio(false);
cin.tie(NULL);
int numberOfTests=1;
//cin>>numberOfTests;
while(numberOfTests--){
solve();
}
}
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---|---|---|---|---|
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