Codeforecs 675E Trains and Statistic 贪心+DP

题意:给出n-1个数,a[i]表示: 从i能直接到达点[i+1,a[i]].p[i][j]为i->j的最短路径.
n<=1e5.i+1<=a[i]<=n问所有p[i][j]的累加和(i=1~n,j=i+1~n)

考虑从i出发[i+1,a[i]]这段只需要1,[a[i]+1,max(a[i+1],a[i+2],a[a[i]])]这段最短距离为2
a[m]=max(a[i+1],a[i+2],a[a[i]],也就是说若移动次数大于1,第一步肯定移动到m点(m点的下一次选择最多)
设dp[i]:从i出发到[i+1,n]的累加和

dp[i]= dp[m]+n-i+1-(a[i]-m). a[m]=max(a[i+1],..a[a[i]])

#include <bits/stdc++.h>using namespace std;typedef long long ll;typedef pair<int,int> ii;const int N=2e5+20;ll n,a[N],dp[N];ii f[N][30];void init(){for(int i=1;i<=n;i++)f[i][0]=ii(a[i],i);for(int j=1;(1<<j)<=n;j++)for(int i=1;i+(1<<j)-1<=n;i++)f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);}ii RMQ(int l,int r){int k=0;while((1<<(k+1))<=r-l+1)k++;return max(f[l][k],f[r-(1<<k)+1][k]);}int main(){while(cin>>n){for(int i=1;i<n;i++)scanf("%I64d",&a[i]);init();ll ans=0;dp[n]=0;for(int i=n-1;i>=1;i--){ii P=RMQ(i+1,a[i]);int m=P.second;dp[i]=dp[m]+n-i-(a[i]-m);ans+=dp[i];}cout<<ans<<endl;}return 0;}