CF675E Trains and Statistic [dp，单调栈]

设 \(f_i = \sum_{j=i+1}^{n} d_{i,j}\)
转移方程是 \(f_i = f_k + (n - k) - (a_i - k) + (k - i)\)
\(n-k\) 是因为我们认定 \(>k\) 的点都会经过 \(k\)
\(a_i-k\) 是因为我们有一部分是重复的，这样算会经过两次，所以要剪掉
\(k-i\) 是因为要加上 \([i+1,k]\) 的点

// powered by c++11
// by Isaunoya
#include <bits/stdc++.h>
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
using namespace std;
using db = double;
using ll = long long;
using uint = unsigned int;
#define Tp template
using pii = pair<int, int>;
#define fir first
#define sec second
Tp<class T> void cmax(T& x, const T& y) {if (x < y) x = y;} Tp<class T> void cmin(T& x, const T& y) {if (x > y) x = y;}
#define all(v) v.begin(), v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
Tp<class T> void sort(vector<T>& v) { sort(all(v)); } Tp<class T> void reverse(vector<T>& v) { reverse(all(v)); }
Tp<class T> void unique(vector<T>& v) { sort(all(v)), v.erase(unique(all(v)), v.end()); }
const int SZ = 1 << 23 | 233;
struct FILEIN { char qwq[SZ], *S = qwq, *T = qwq, ch;
#ifdef __WIN64
#define GETC getchar
#else
  char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq, 1, SZ, stdin), S == T) ? EOF : *S++; }
#endif
  FILEIN& operator>>(char& c) {while (isspace(c = GETC()));return *this;}
  FILEIN& operator>>(string& s) {while (isspace(ch = GETC())); s = ch;while (!isspace(ch = GETC())) s += ch;return *this;}
  Tp<class T> void read(T& x) { bool sign = 0;while ((ch = GETC()) < 48) sign ^= (ch == 45); x = (ch ^ 48);
    while ((ch = GETC()) > 47) x = (x << 1) + (x << 3) + (ch ^ 48); x = sign ? -x : x;
  }FILEIN& operator>>(int& x) { return read(x), *this; } FILEIN& operator>>(ll& x) { return read(x), *this; }
} in;
struct FILEOUT {const static int LIMIT = 1 << 22 ;char quq[SZ], ST[233];int sz, O;
  ~FILEOUT() { flush() ; }void flush() {fwrite(quq, 1, O, stdout); fflush(stdout);O = 0;}
  FILEOUT& operator<<(char c) {return quq[O++] = c, *this;}
  FILEOUT& operator<<(string str) {if (O > LIMIT) flush();for (char c : str) quq[O++] = c;return *this;}
  Tp<class T> void write(T x) {if (O > LIMIT) flush();if (x < 0) {quq[O++] = 45;x = -x;}
		do {ST[++sz] = x % 10 ^ 48;x /= 10;} while (x);while (sz) quq[O++] = ST[sz--];
  }FILEOUT& operator<<(int x) { return write(x), *this; } FILEOUT& operator<<(ll x) { return write(x), *this; }
} out;
#define int long long
signed main() {
  // code begin.
	int n ;
	in >> n ;
	vector < int > a(n , 0) , vec , f(n , 0) ;
	for(int i = 1 ; i < n ; i ++)
		in >> a[i] ;
	int ans = (f[n - 1] = 1) ;
	vec.push_back(n - 1) ;
	for(int i = n - 2 ; i ; i --) {
		int k = * (upper_bound(vec.rbegin() , vec.rend() , a[i]) - 1) ;
		f[i] = f[k] + (n - k) - (a[i] - k) + (k - i) ;
		ans += f[i] ;
		while(! vec.empty() && a[i] >= a[vec.back()])
			vec.pop_back() ;
		vec.push_back(i) ;
	}
	out << ans << '\n' ;
	return 0;
  // code end.
}