洛谷传送门

题目描述

解析

利用倍增，设计dp慢慢敲即可。。。
注意距离累加在一起会爆int，需要ll
特判条件非常之复杂。。。
心力交瘁，就酱了

代码

#include <bits/stdc++.h>
using namespace std;
#define ll long long
typedef pair<ll,ll> pr;
const int N = 3e5 + 100;
int n,m;
struct node{int id,h;bool operator < (const node y)const{return h<y.h;}
}p[N];
int pos[N];
int l[N],r[N],h[N];
bool ok(int x){return x>=1&&x<=n;}
ll jl(int x,int y){return abs(p[x].h-p[y].h);}
struct node2{ll dist,id;bool operator < (const node2 y)const{if(dist!=y.dist) return dist<y.dist;else return h[id]<h[y.id];}
};
node2 fir[N],sec[N];
void del(int x){r[l[x]]=r[x];l[r[x]]=l[x];return;
}
int pl[N][30];
ll dis[3][N][30];
int x0;
void solve(){sort(p+1,p+1+n);for(int i=1;i<=n;i++) pos[p[i].id]=i;for(int i=1;i<=n;i++){l[i]=i-1;r[i]=i+1;}r[0]=1;l[n+1]=n;for(int i=1;i<=n;i++){int pl=pos[i];int l1=0,l2=0,r1=0,r2=0;int num=0;node2 coi[5];if(ok(l[pl])) l1=l[pl],coi[++num]=(node2){jl(pl,l1),p[l1].id};if(l1&&ok(l[l1])) l2=l[l1],coi[++num]=(node2){jl(pl,l2),p[l2].id};if(ok(r[pl])) r1=r[pl],coi[++num]=(node2){jl(pl,r1),p[r1].id};if(r1&&ok(r[r1])) r2=r[r1],coi[++num]=(node2){jl(pl,r2),p[r2].id};sort(coi+1,coi+1+num);if(num>=1) fir[i]=coi[1];if(num>=2) sec[i]=coi[2];del(pl);//printf("i=%d fir=%d sec=%d\n",i,fir[i].id,sec[i].id);}for(int i=1;i<=n;i++){if(sec[i].id) pl[i][0]=sec[i].id;else pl[i][0]=i;}for(int i=1;i<=n;i++){if(pl[i][0]!=i&&fir[pl[i][0]].id) pl[i][1]=fir[pl[i][0]].id;else pl[i][1]=pl[i][0];}for(int i=1;i<=n;i++){if(sec[i].id) dis[1][i][0]=sec[i].dist;else dis[1][i][0]=0;dis[2][i][0]=0;}for(int i=1;i<=n;i++){dis[1][i][1]=dis[1][i][0];if(pl[i][0]!=i&&fir[pl[i][0]].id)dis[2][i][1]=fir[pl[i][0]].dist;else dis[2][i][1]=0;}for(int k=2;(1<<k)<=n;k++){for(int i=1;i<=n;i++){pl[i][k]=pl[pl[i][k-1]][k-1];dis[1][i][k]=dis[1][i][k-1]+dis[1][pl[i][k-1]][k-1];dis[2][i][k]=dis[2][i][k-1]+dis[2][pl[i][k-1]][k-1];}}
}
pair<ll,ll> find(int st,int x){
//	printf("ask: st=%d x=%d\n",st,x);ll disa=0,disb=0,tot=0;int ppl=st;for(int k=20;k>=0;k--){if((1<<k)>n) continue;ll suma=dis[1][ppl][k],sumb=dis[2][ppl][k];if(suma+sumb+tot>x) continue;tot+=suma+sumb;disa+=suma;disb+=sumb;ppl=pl[ppl][k];}
//	printf("  pl=%d disa=%d disb=%d\n",ppl,disa,disb);return make_pair(disa,disb);
}
void test(){for(int k=0;k<=3;k++){for(int i=1;i<=n;i++){printf("i=%d k=%d pl=%d dis1=%d dis2=%d\n",i,k,pl[i][k],dis[1][i][k],dis[2][i][k]);}}return ;
}
int main() {scanf("%lld",&n);for(int i=1;i<=n;i++){scanf("%lld",&p[i].h);h[i]=p[i].h;p[i].id=i;}solve();//test();scanf("%lld",&x0);double mn=2e16,temp;int anspl;for(int i=1;i<=n;i++){pr o=find(i,x0);//printf("st=%d disa=%d disb=%d\n",i,o.first,o.second);temp= o.second==0?2e15:1.0*o.first/o.second;if(temp<mn){mn=temp;anspl=i;}else if(temp==mn&&p[pos[anspl]].h<p[pos[i]].h) anspl=i;}printf("%lld\n",anspl);scanf("%lld",&m);int s,x;for(int i=1;i<=m;i++){scanf("%d%d",&s,&x);pr o=find(s,x);printf("%lld %lld\n",o.first,o.second);}return 0;
}
/*
10 
4 5 6 1 2 3 7 8 9 10 
7 
10 
1 7 
2 7 
3 7 
4 7 
5 7 
6 7 
7 7 
8 7 
9 7 
10 7
*/