Luogu1081 NOIP2012 开车旅行 倍增

题目传送门

为什么NOIP的题目都这么长qwq

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话说2012的D1T3和D2T3都是大火题啊qwq

预处理神题

对于这种跳跳跳的题目考虑使用倍增优化枚举。先预处理某个点之后距离最小和次小的城市，然后倍增预处理一大堆东西。设$f_i$表示从$A$开始开$2^i$次车到达的地点，$g_i$表示从$B$开始开$2^i$次车到达的地点，$k_{i,0/1}$表示从$A$开始开$2^i$次车$A$或$B$经过的路程，$l_{i,0/1}$表示从$B$开始开$2^i$次车$A$或$B$经过的路程。注意边界与越界情况的处理。

接下来就可以直接拼凑法跳了，时间复杂度$O((n+q)logn)$

 #include<bits/stdc++.h>
 #define MAXN 100001
 using namespace std;
 inline int read(){
      , k = ;
     char c = getchar();
     while(!isdigit(c) && c != '-')  c = getchar();
     if(c == '-')    k = -k , c = getchar();
     ) + a + (c ^ ') , c = getchar();
     return a * k;
 }
 ][] , lb[MAXN][][];
 ][] , h[MAXN] , N;
 set < pair < int , int > > s;
 inline  ? a : -a;}
 pair < int , int > cul(int ss , int x){
     pair <  , );
     ) ; i >=  ; i--)
         ] && (] + lb[ss][i][] <= x){
             t.first += la[ss][i][];
             t.second += lb[ss][i][];
             ss = f[ss][i][];
         }
     return t;
 }
 int main(){
     memset(la ,  ,  , sizeof(lb));
     N = read();
      ; i <= N ; i++)   h[i] = read();
     s.insert(make_pair(h[N] , N));
     s.insert(make_pair(h[N - ] , N - ));
     )   f[N - ][][] = N;
      ; i ; i--){
         set < pair < int , int > > :: iterator t = s.insert(make_pair(h[i] , i)).first;
         if(t == s.begin()){
             f[i][][] = (++s.begin())->second;
             f[i][][] = (++++s.begin())->second;
             continue;
         }
         set < pair < int , int > > :: iterator t1 = --t , t2 = ++++t;
         if(t2 == s.end()){
             f[i][][] = t1->second;
             f[i][][] = (--t1)->second;
         }
         else{
             if(h[i] - t1->first <= t2->first - h[i]){
                 f[i][][] = t1->second;
                 if(t1 == s.begin()){
                     f[i][][] = t2->second;
                     continue;
                 }
                 --t1;
             }
             else{
                 f[i][][] = t2->second;
                 if(++t2 == s.end()){
                     f[i][][] = t1->second;
                     continue;
                 }
             }
             ][] = t1->second;
             ][] = t2->second;
         }
     }
      ; i <= N ; i++){
         ][])  la[i][][] = abs(h[i] - h[f[i][][]]);
         ][])  lb[i][][] = abs(h[i] - h[f[i][][]]);
         la[i][][] = lb[i][][] = ;
     }
      ; i < N ; i++){
         f[i][][] = f[f[i][][]][][];
         f[i][][] = f[f[i][][]][][];
         la[i][][] = la[i][][];
         lb[i][][] = lb[f[i][][]][][];
         la[i][][] = la[f[i][][]][][];
         lb[i][][] = lb[i][][];
     }
      ;  << i < N ; i++)
          ; j <= N - ( << i) ; j++){
             f[j][i][] = f[f[j][i - ][]][i - ][];
             f[j][i][] = f[f[j][i - ][]][i - ][];
             la[j][i][] = la[j][i - ][] + la[f[j][i - ][]][i - ][];
             lb[j][i][] = lb[j][i - ][] + lb[f[j][i - ][]][i - ][];
             la[j][i][] = la[j][i - ][] + la[f[j][i - ][]][i - ][];
             lb[j][i][] = lb[j][i - ][] + lb[f[j][i - ][]][i - ][];
         }
     ;
     pair <  , X0);
      ; i <= N ; i++){
         pair < int , int > t = cul(i , X0);
         ){
             maxN = t;
             dir = i;
         }
          && t.second ==  && h[dir] < h[i]){
             maxN = t;
             dir = i;
         }
         else    if(t.second && ((long long)maxN.first * t.second > (long long)maxN.second * t.first || maxN.first * t.second == maxN.second * t.first && h[dir] < h[i])){
             maxN = t;
             dir = i;
         }
     }
     printf("%d\n" , dir);
     for(X0 = read() ; X0 ; X0--){
         int a = read() , b = read();
         pair < int , int > t = cul(a , b);
         printf("%d %d\n" , t.first , t.second);
     }
     ;
 }