題目傳送門

為什麽NOIP的題目都這麽長qwq

話說2012的D1T3和D2T3都是大火題啊qwq

預處理神題

對於這種跳跳跳的題目考慮使用倍增優化枚舉。先預處理某個點之後距離最小和次小的城市，然後倍增預處理一大堆東西。設$f_i$表示從$A$開始開$2^i$次車到達的地點，$g_i$表示從$B$開始開$2^i$次車到達的地點，$k_{i,0/1}$表示從$A$開始開$2^i$次車$A$或$B$經過的路程，$l_{i,0/1}$表示從$B$開始開$2^i$次車$A$或$B$經過的路程。註意邊界與越界情況的處理。

接下來就可以直接拼湊法跳了，時間復雜度$O((n+q)logn)$

  1 #include<bits/stdc++.h>
  2
 
 #define MAXN 100001
  3 using namespace std;
  4 inline int read(){
  5     int a = 0 , k = 1;
  6     char c = getchar();
  7     while(!isdigit(c) && c != ‘-‘)  c = getchar();
  8     if(c == ‘-‘)    k = -k , c = getchar();
  9     while(isdigit(c))   a += (a << 3) + a + (c ^ ‘0‘) , c = getchar();
 
 10     return a * k;
 11 }
 12 long long la[MAXN][17][2] , lb[MAXN][17][2];
 13 int f[MAXN][17][2] , h[MAXN] , N;
 14 set < pair < int , int > > s;
 15 inline int abs(int a){return a > 0 ? a : -a;}
 16 pair < int , int > cul(int ss , int x){
 17     pair < int , int > t(0 , 0
 
);
 18     for(int i = log(N) / log(2) ; i >= 0 ; i--)
 19         if(f[ss][i][0] && (long long)t.first + t.second + la[ss][i][0] + lb[ss][i][0] <= x){
 20             t.first += la[ss][i][0];
 21             t.second += lb[ss][i][0];
 22             ss = f[ss][i][0];
 23         }
 24     return t;
 25 }
 26 int main(){
 27     memset(la , 1 , sizeof(la));    memset(lb , 1 , sizeof(lb)); 
 28     N = read();
 29     for(int i = 1 ; i <= N ; i++)   h[i] = read();
 30     s.insert(make_pair(h[N] , N));
 31     s.insert(make_pair(h[N - 1] , N - 1));
 32     if(N - 1)   f[N - 1][0][1] = N;
 33     for(int i = N - 2 ; i ; i--){
 34         set < pair < int , int > > :: iterator t = s.insert(make_pair(h[i] , i)).first;
 35         if(t == s.begin()){
 36             f[i][0][1] = (++s.begin())->second;
 37             f[i][0][0] = (++++s.begin())->second;
 38             continue;
 39         }
 40         set < pair < int , int > > :: iterator t1 = --t , t2 = ++++t;
 41         if(t2 == s.end()){
 42             f[i][0][1] = t1->second;
 43             f[i][0][0] = (--t1)->second;
 44         }
 45         else{
 46             if(h[i] - t1->first <= t2->first - h[i]){
 47                 f[i][0][1] = t1->second;
 48                 if(t1 == s.begin()){
 49                     f[i][0][0] = t2->second;
 50                     continue;
 51                 }
 52                 --t1;
 53             }
 54             else{
 55                 f[i][0][1] = t2->second;
 56                 if(++t2 == s.end()){
 57                     f[i][0][0] = t1->second;
 58                     continue;
 59                 }
 60             }
 61             if(h[i] - t1->first <= t2->first - h[i])    f[i][0][0] = t1->second;
 62             else    f[i][0][0] = t2->second;
 63         }
 64     }
 65     for(int i = 1 ; i <= N ; i++){
 66         if(f[i][0][0])  la[i][0][0] = abs(h[i] - h[f[i][0][0]]);
 67         if(f[i][0][1])  lb[i][0][1] = abs(h[i] - h[f[i][0][1]]);
 68         la[i][0][1] = lb[i][0][0] = 0;
 69     }
 70     for(int i = 1 ; i < N ; i++){
 71         f[i][1][0] = f[f[i][0][0]][0][1];
 72         f[i][1][1] = f[f[i][0][1]][0][0];
 73         la[i][1][0] = la[i][0][0];
 74         lb[i][1][0] = lb[f[i][0][0]][0][1];
 75         la[i][1][1] = la[f[i][0][1]][0][0];
 76         lb[i][1][1] = lb[i][0][1];
 77     }
 78     for(int i = 2 ; 1 << i < N ; i++)
 79         for(int j = 1 ; j <= N - (1 << i) ; j++){
 80             f[j][i][0] = f[f[j][i - 1][0]][i - 1][0];
 81             f[j][i][1] = f[f[j][i - 1][1]][i - 1][1];
 82             la[j][i][0] = la[j][i - 1][0] + la[f[j][i - 1][0]][i - 1][0];
 83             lb[j][i][0] = lb[j][i - 1][0] + lb[f[j][i - 1][0]][i - 1][0];
 84             la[j][i][1] = la[j][i - 1][1] + la[f[j][i - 1][1]][i - 1][1];
 85             lb[j][i][1] = lb[j][i - 1][1] + lb[f[j][i - 1][1]][i - 1][1];
 86         }
 87     int X0 = read() , dir = 1;
 88     pair < int , int > maxN = cul(1 , X0);
 89     for(int i = 2 ; i <= N ; i++){
 90         pair < int , int > t = cul(i , X0);
 91         if(t.second && maxN.second == 0){
 92             maxN = t;
 93             dir = i;
 94         }
 95         else    if(maxN.second == 0 && t.second == 0 && h[dir] < h[i]){
 96             maxN = t;
 97             dir = i;
 98         }
 99         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])){
100             maxN = t;
101             dir = i;
102         }
103     }
104     printf("%d\n" , dir);
105     for(X0 = read() ; X0 ; X0--){
106         int a = read() , b = read();
107         pair < int , int > t = cul(a , b);
108         printf("%d %d\n" , t.first , t.second);
109     }
110     return 0;
111 }

Luogu1081 NOIP2012D1T3 開車旅行 倍增