abc435_f
不理解为什么都不会，其实挺简单的。
思路
首先，猫一开始在最高点，考虑我下一步有意义的操作只有撤掉最高的塔（若撤掉别的塔，对猫没有任何影响，只会减小答案）。
考虑撤掉 \([l,r]\) 最高塔（位置\(P\)）后的两种情况
- 猫跳到左边最高塔（位置\(p_l\)），这步跳跃对答案贡献 \(del_l=(P-p_l)\) 。
- 猫跳到右边最高塔（位置\(p_r\)），这步跳跃对答案贡献 \(del_r=(p_r-P)\) 。
那么此时的答案 \(solve(l,r)\) 就是 \(\max(solve(l,P)+del_l, solve(P,r)+del_r)\) 从 \(solve(1,n)\) 开始递归即可。
那么我们只需要快速处理任意一段区间的最大值即可，考虑倍增维护，于是复杂度为预处理 \(O(n\log(n))\) ，计算 \(O(n)\) 显然正确。
Code
// By wnn
#include<bits/stdc++.h>
// simple name
#define pii pair<int,int>
#define pll pair<long long, long long>
#define pque priority_queue
#define x1 x_1
#define y1 y_1
#define fir first
#define sec second
#define pb push_back
#define myfreopen freopen(".in", "r", stdin),freopen(".out", "w", stdout)
// function
#define ls(x) (x << 1)
#define rs(x) ((x << 1) | 1)
#define mid(l, r) ((l + r) >> 1)
#define debug(x) cerr << x << endl
#define dist(x, y, x2, y2) sqrt((x - x2) * (x - x2) + (y - y2) * (y - y2))
#define WA cerr << "Wrong Answer" << endl
#define init_inf32(x) memset(x, 0x3f, sizeof(x))
#define init_inf64(x) memset(x, 0x3fll, sizeof(x))
#define init_0(x) memset(x, 0, sizeof(x))
#define Dec(x) fixed << setprecision(x)
// val
#define eps 1e-9
#define inf32 0x3f3f3f3f
#define inf64 0x3f3f3f3f3f3f3f3fll
#define mod1 (int)(1e9 + 7)
#define mod2 998244353
#define PI acos(-1.0)
// god
#define int long long
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// Def important function
inline void init();
inline void ever_init();
inline void solve();
// Init rnd()
mt19937 rnd(time(0) ^ clock());
// Constants
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};
const int N = 2e5 + 5;int n, p[N], pos[N];
int f[N][20];
inline void init1(){for(int i = 1; i <= n; i++){f[i][0] = p[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]);}
}
inline int cal(int l, int r){int k = (log2(r - l + 1));return max(f[l][k], f[r - (1 << k) + 1][k]);
}
inline int dfs(int l, int r){if(l >= r) return 0;int maxn = cal(l, r), mp = pos[maxn];int len1 = mp - pos[cal(l, mp - 1)], len2 = pos[cal(mp + 1, r)] - mp;return max(dfs(l, mp - 1) + len1, dfs(mp + 1, r) + len2);
}inline void init(){cin >> n;for(int i = 1; i <= n; pos[p[i]] = i, i++) cin >> p[i];init1();cout << dfs(1, n);
}
inline void ever_init(){}
inline void solve(){}signed main(){
// myfreopen;ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);init();int T = 1;
// cin >> T;while(T--){ever_init();solve();}return 0;
}
/*things to check:
* Will it MLE?
* Is array big enough?
* Do you need long long?
* Is inf big enough?
* max or min?
* Yes,No or YES,NO?
* Is there anything extra to output?
* Did you Countershoot?
* Have you measured the limit data?
* More measurements should be cleared!!!
*/