codeforces 675E E. Trains and Statistic(线段树+dp)

题目链接：

E. Trains and Statistic

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya commutes by train every day. There are n train stations in the city, and at the i-th station it's possible to buy only tickets to stations from i + 1 to ai inclusive. No tickets are sold at the last station.

Let ρi, j be the minimum number of tickets one needs to buy in order to get from stations i to station j. As Vasya is fond of different useless statistic he asks you to compute the sum of all values ρi, j among all pairs 1 ≤ i < j ≤ n.

 

Input

 

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of stations.

The second line contains n - 1 integer ai (i + 1 ≤ ai ≤ n), the i-th of them means that at the i-th station one may buy tickets to each station from i + 1 to ai inclusive.

 

Output

 

Print the sum of ρi, j among all pairs of 1 ≤ i < j ≤ n.

 

Examples

 

input

4
4 4 4

output

6

input

5
2 3 5 5

output

17



题意：

a[i]表示从第i个车站可以一张票到第[i+1,a[i]]这些车站;
p[i][j]表示从第i个车站到第j个车站的最少的票数,现在要求∑dp[i][j](1<=i<=n,i<j<=n);


思路：

dp[i]=∑p[i][j](i<j<=n);
转移方程为dp[i]=dp[temp]+(n-i)-(a[i]-temp);
其中temp表示[i+1，a[i]]中a[temp]最大的位置;

AC代码：

#include <bits/stdc++.h>
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int N=1e5+25;
int n,a[N];
LL dp[N];
struct Tree
{
    int l,r,ans;
}tr[4*N];
void pushup(int o)
{
    if(a[tr[2*o].ans]>a[tr[2*o+1].ans])tr[o].ans=tr[2*o].ans;
    else tr[o].ans=tr[2*o+1].ans;
}
void build(int o,int L,int R)
{
    tr[o].l=L;
    tr[o].r=R;
    if(L==R)
    {
        tr[o].ans=L;
        return ;
    }
    int mid=(L+R)>>1;
    build(2*o,L,mid);
    build(2*o+1,mid+1,R);
    pushup(o);
}
int query(int o,int L,int R)
{
    if(L<=tr[o].l&&R>=tr[o].r)return tr[o].ans;
    int mid=(tr[o].l+tr[o].r)>>1;
    if(R<=mid)return query(2*o,L,R);
    else if(L>mid)return query(2*o+1,L,R);
    else
    {
        int fl=query(2*o,L,mid),fr=query(2*o+1,mid+1,R);
        if(a[fl]>a[fr])return fl;
        else return fr;
    }
}
int main()
{
    scanf("%d",&n);
    Riep(n-1)scanf("%d",&a[i]);
    a[n]=n-1;
    build(1,1,n);
    LL ans=0;
    dp[n]=0;
    for(int i=n-1;i>0;i--)
    {
        int temp=query(1,i+1,a[i]);
        dp[i]=dp[temp]+(LL)(n-i-(a[i]-temp));
        ans+=dp[i];
    }
    cout<<ans<<"\n";

    return 0;
}