# codeforces675e Trains and Statistic RMQ optimization dp+ greedy

Source: Internet
Author: User

Test instructions

A[i] means to buy a ticket from the station I can sit at any of the [I+1,a[i]] Stations
P[I][J] Indicates the minimum number of votes from station I to J Station, ∑p[i][j] (1<=i<=n,i<j<=n)

Ideas:

Construction dp[i]=∑p[i][j](i<j≤n)

greedy to think, every time from station I to buy a ticket to the first station (in [I+1,a[i], A[m] the largest), and then the choice of car is the best, because each time can make the same number of votes to go the furthest . Unless the destination is within [i+1,a[i]] This interval, do not need to go far.

This can be classified as a discussion, assuming that all routes are first bought M station, then Dp[i] can be linearly represented by dp[m], and [i+1,m] in the interval wherever the seat is a ticket, no effect on the result, and because [m,a[i]] This zone only need to buy tickets at the I station can be sat over, And according to the previous idea requires 2 tickets, so to reduce the number of extra votes in this interval.

So,dp[i] = Dp[m] + (n-i) {All routes first buy M station}-(A[i]-m) {Multiple ticket purchase interval}

The problem solved the spread of flowers ~

1#include <bits/stdc++.h>2 using namespacestd;3typedefLong LongLL;4 Const intMAXN =100005;5 inta[maxn],maxsum[maxn][ -],n;6vector<int>V[MAXN];7 LL DP[MAXN];8 voidRMQ (intnum) {9      for(intj=1; j< -; J + +)Ten          for(intI=1; i<=num; i++) One             if(i + (1<< j)-1<=num) AMAXSUM[I][J] = max (maxsum[i][j-1], maxsum[i+ (1<< (J-1))][j-1]); - } - intMain () the { -scanf"%d",&n); -      for(intI=1; i<n; i++) { -scanf"%d",&a[i]); + V[a[i]].push_back (i); -     } +      for(intI=2; i<=n; i++) { A v[i].push_back (n); at sort (V[i].begin (), V[i].end ()); -     } -      for(intI=1; i<=n; i++) -maxsum[i][0] =A[i]; - RMQ (n); -LL ans =0; inDp[n] =0; -      for(inti=n-1; I>0; i--) { to         intx = i, y =A[i]; +         intT = (int) (Log (Y-x +1.0)/log (2.0)); -         intMAXT = Max (Maxsum[x][t], maxsum[y-(1&LT;&LT;T) +1][t]); the         intm = Lower_bound (V[maxt].begin (), V[maxt].end (), i +1) -V[maxt].begin (); *m =V[maxt][m]; \$Dp[i] = dp[m] + n-i-(A[i]-m);Panax NotoginsengAns + =Dp[i]; -     } theprintf"%i64d\n", ans); +     return 0; A}
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codeforces675e Trains and Statistic RMQ optimization dp+ greedy

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