【34.54%】【codeforces 675E】Trains and Statistic

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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

Note

In the first sample it’s possible to get from any station to any other (with greater index) using only one ticket. The total number of pairs is 6, so the answer is also 6.

Consider the second sample:

ρ1, 2 = 1

ρ1, 3 = 2

ρ1, 4 = 3

ρ1, 5 = 3

ρ2, 3 = 1

ρ2, 4 = 2

ρ2, 5 = 2

ρ3, 4 = 1

ρ3, 5 = 1

ρ4, 5 = 1

Thus the answer equals 1 + 2 + 3 + 3 + 1 + 2 + 2 + 1 + 1 + 1 = 17.

【题解】



设dp[i]为i到i+1..n这些点的”总共”需要邮票的张数;

转移方程

dp[i] = dp[pos]+n-i-(a[i]-pos);

这里的pos,是(i+1..a[i])这个区间里面的a的值最大的下标;

表示的是从i开始往后转移的方式;

首先

从i->(i+1..a[i])都是只要一张票的。(这肯定是最优的)。所以不用从i到pos再转移到(pos+1..a[i]);

直接到pos+1..a[i]就可以了；

而a[i]之后的位置。则如果要到a[i]..a[pos]这段。则需要从i到pos再转移到。如下图。



上图中0->1、2、3、4、5、6这些都只要买一张票;

而dp[pos]包含了从pos->4、5的两张票；

如果再加上了dp[pos],则会重复;

因此要减去(a[i]-pos);

n-i+a[i]-pos

对应:

pos->a[i]后面的点;

然后从i连若干条边到pos;

因为有n-i+a[i]-pos个点。

所以总共要加n-i+a[i]-pos条边;

为什么选a最大的pos;

因为a[pos]是最大的。所以从pos可以走到更远的地方



如图，如果选择的是4。它最远能到8。则如果要到9就的再从8再买一张票到9；

而如果选择pos,则可以直接买一张票到9；这就节省了一张。

画图多理解下吧。

最后累加dp[1..n];

dp[n] = 0;

一开始dp[n-1]=1;

从n-2开始往前处理

获取某个区间内的最大值用ST算法就好(线段树is ok)

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#define LL long long

using namespace std;

const int MAXN = 200000;
const int MAX = 17;

int n, a[MAXN],rmq[MAXN][MAX+3],pre[MAX+3];
int need[MAXN];
LL dp[MAXN],ans = 0;

void input(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)) t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

int main()
{
    //freopen("F:\\rush.txt", "r", stdin);
    scanf("%d", &n);
    for (int i = 1; i <= n-1; i++)
        scanf("%d", &a[i]),rmq[i][0] = i;

    pre[0] = 1;
    for (int i = 1; i <= MAX; i++)
        pre[i] = pre[i - 1] * 2;

    int now = 1;
    need[1] = 0;
    for (int i = 2; i <= n; i++)
        if (i == pre[now])
            need[i] = need[i - 1] + 1,now++;
        else
            need[i] = need[i - 1];

    for (int l = 1;pre[l]<=n;l++)
        for (int i = 1; i + pre[l]-1<= n; i++)
        {
            int left = rmq[i][l-1];
            int right = rmq[i + pre[l - 1]][l-1];
            if (a[left] > a[right])
                rmq[i][l] = left;
            else
                    rmq[i][l] = right;
        }

    dp[n-1] = 1;
    ans = 1;
    for (int i = n - 2; i >= 1; i--)
    {
        int xy = need[a[i] - i+1];
        int left = rmq[i][xy], right = rmq[a[i] - pre[xy] + 1][xy];
        int pos;
        if (a[left] > a[right])
            pos = left;
        else
            pos = right;
        dp[i] = dp[pos] + n - i - (a[i] - pos);
        ans += dp[i];
    }

    printf("%I64d\n", ans);
    return 0;
}