Thoughts:
premise sth.\(nxt1_i\) denotes the distance to the nearest city on the right (\(id1_i\) (for numbering), so that\(nxt2_i\) denotes the second closest city number on the right (\(id2_i\) (for numbering); it is possible to useset Find the nearest city to this\(4\) cities (two in the front and two in the back).
Definition:
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\(f_{i,j}\) denote\(i\) leave at 1:00 p.m. and go\(2^j\) The wheel arrives at the last position.
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\(dp1_{i,j}\) denote\(i\) leave at 1:00 p.m. and go\(2^j\) The last A of the round traveled the distance.
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\(dp2_{i,j}\) denote\(i\) leave at 1:00 p.m. and go\(2^j\) Round the distance traveled by the last B.
Initialization:
The state transfer equation is:
At this point for inquiries\(1\) and inquiries\(2\)：
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Essentially, it's finding out that after traveling from each city\(A\) Distance traveled vs.\(B\) Walking distance.
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Then consider greed from high to low, i.e. set the current jump to the\(s\) point, if\(dp1_{s,i} + dp2_{s,i} \le x\)can be obtained from the\(s\) jump to\(f_{s,i}\)The need to make\(x \gets x - (dp1_{s,i} + dp2_{s,i})\)and then continue traversing\(i-1\) Bit.
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Since A drove first, A may be able to drive again at the end of the last round, requiring a special judgment.
The time complexity is\(O((N+Q) \log N)\)。
Full Code:
#include<bits/stdc++.h>
#define Add(x,y) (x+y>=mod)?(x+y-mod):(x+y)
#define lowbit(x) x&(-x)
#define pi pair<ll,ll>
#define pii pair<ll,pair<ll,ll>>
#define iip pair<pair<ll,ll>,ll>
#define ppii pair<pair<ll,ll>,pair<ll,ll>>
#define fi first
#define se second
#define full(l,r,x) for(auto it=l;it!=r;it++) (*it)=x
#define Full(a) memset(a,0,sizeof(a))
#define open(s1,s2) freopen(s1,"r",stdin),freopen(s2,"w",stdout);
using namespace std;
typedef double db;
typedef unsigned long long ull;
typedef long long ll;
bool Begin;
const ll N=1e5+10,M=18,INF=1e18;
inline ll read(){
ll x=0,f=1;
char c=getchar();
while(c<'0'||c>'9'){
if(c=='-')
f=-1;
c=getchar();
}
while(c>='0'&&c<='9'){
x=(x<<1)+(x<<3)+(c^48);
c=getchar();
}
return x*f;
}
inline void write(ll x){
if(x<0){
putchar('-');
x=-x;
}
if(x>9)
write(x/10);
putchar(x%10+'0');
}
struct Node{
ll a,b;
ll id;
bool operator<(const Node&rhs)const{
if(a!=)
return a<;
return b<;
}
};
ll n,m,s,x,x0,s0,cnt,dis1,dis2,disa,disb;
ll a[N],nxt1[N],nxt2[N],id1[N],id2[N];
ll f[N][M],dp1[N][M],dp2[N][M];
Node h[N];
vector<Node> V;
multiset<Node> S;
void solve(ll s,ll x){
dis1=dis2=0;
for(int i=M-1;i>=0;i--){
if(dp1[s][i]+dp2[s][i]<=x&&f[s][i]){
dis1+=dp1[s][i],dis2+=dp2[s][i];
x-=dp1[s][i]+dp2[s][i];
s=f[s][i];
}
}
if(nxt2[s]<=x)
dis1+=nxt2[s];
}
bool End;
int main(){
n=read();
for(int i=1;i<=n;i++){
a[i]=read();
h[i]={a[i],a[i],i};
}
({INF,INF,0});
({INF-1,INF-1,0});
({-INF,-INF,0});
({-INF+1,-INF+1,0});
for(int i=n;i>=1;i--){
();
V.push_back(*--S.lower_bound(h[i]));
V.push_back(*--S.lower_bound(V[0]));
V.push_back(*S.upper_bound(h[i]));
V.push_back(*S.upper_bound(V[2]));
for(auto &v:V)
=abs(h[i].);
sort((),());
nxt1[i]=V[0].a,nxt2[i]=V[1].a;
id1[i]=V[0].id,id2[i]=V[1].id;
cerr<<id1[i]<<' '<<id2[i]<<'\n';
(h[i]);
}
for(int i=1;i<=n;i++){
f[i][0]=id1[id2[i]];
dp1[i][0]=nxt2[i];
dp2[i][0]=nxt1[id2[i]];
}
for(int j=1;j<M;j++){
for(int i=n;i>=1;i--){
f[i][j]=f[f[i][j-1]][j-1];
dp1[i][j]=dp1[i][j-1]+dp1[f[i][j-1]][j-1];
dp2[i][j]=dp2[i][j-1]+dp2[f[i][j-1]][j-1];
}
}
x0=read();
for(int i=1;i<=n;i++){
solve(i,x0);
if(dis2&&(!s0||disa*dis2>disb*dis1)){
s0=i;
disa=dis1,disb=dis2;
}
}
write(s0);
putchar('\n');
m=read();
while(m--){
s=read(),x=read();
solve(s,x);
write(dis1);
putchar(' ');
write(dis2);
putchar('\n');
}
cerr<<'\n'<<abs(&Begin-&End)/1048576<<"MB";
return 0;
}