[题目链接]([P2501 HAOI2006] 数字序列 - 洛谷 | 计算机科学教育新生态)
首先是第一问，直接求不好求，我们们考虑求不用更改的数量，发现有这个性质，如果，a[i] - a[j] < abs(j - i)两个数的差值能满足他们之间有足够多数的情况，例如1 4 5 3，取1 和 3，那么就有2 < 3, 中间的4 和 5 怎么改也不会单调上升
所以， 经过移项得a[i] - i < a[j] - j 即可，这就具有了，求最长的上升a[i] - i 即可，用b数组存下来是求最大上身子序列
这是第一问
第二问我们假如以i为最后一位的子序列的前一个状态时pre[i] 我们要把之间的数全都修改一遍，那就得让左边某一部分等于b[from],另一部分等于b[i]，枚举这个分区即可求得从from到i这区间变为单调递增的最小花费，总花费加上from之前的就行
// Problem: P2501 [HAOI2006] 数字序列
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P2501
// Memory Limit: 125 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)//by codeforcer ——
// ____ _ _ _ _ _ _ ____ _
// / ___|| | | || | | || | | | |___ \ | |
//| | | |_| || |_| || |_| | __) | | |
//| | | _ || _ || _ | |__ < | |
//| |___ | | | || | | || | | | ___) | | |
// \____||_| |_||_| |_||_| |_| |____ / |_|#include<bits/stdc++.h>
using namespace std;typedef int E;
typedef long long LL;
typedef pair<int, int> PII;
typedef tuple<int, int, int> PIII;
typedef tuple<LL, LL, LL> PLLL;
typedef pair<long long, long long> PLL;
typedef unsigned long long ULL;#define endl '\n'
#define vec vector
#define pb push_back
#define pob pop_back
#define fir first
#define sec second
#define maxINT 0x3f3f3f3f
#define maxLL 0x3f3f3f3f3f3f3f3fLL
#define umap unordered_map
#define uset unordered_set
#define maxheap priority_queue<E, vector<E>, less<E>>
#define minheap priority_queue<E, vector<E>, greater<E>>#define prvec(a) \for (int i = 0; i < (a).size(); i++) { \cout << (a)[i] << " "; \} \cout << endl;#define debugvec(a, i, n) \cout << #a << ": "; \for (int k = (i); k <= (n); k++) { \cout << (a)[k] << " "; \} \cout << endl; LL gcd(LL a, LL b) { return (b) ? gcd(b, a % b) : a; }
LL exgcd(LL a, LL b, LL &x, LL &y) {if (b == 0) { x = 1, y = 0; return a; }LL gcd = exgcd(b, a % b, y, x);y -= a / b * x;return gcd;}
LL qmi(LL a, LL b, LL mod) {LL res = 1;while (b) {if (b & 1) res = res * a % mod;a = a * a % mod;b >>= 1;}return res;}const int N = 3e4 + 5e3;
long long top[N],f[N],g[N];
long long pre[N],sub[N],b[N];
vector<int> ed[N];void solve() {int n;cin>>n;vec<long long> a(n + 3);for(int i = 1;i <= n;i ++){cin>>a[i];b[i] = a[i] - i;}b[0] = -(1e9+2093),b[n + 1] = 1e9 + 2330;int cnt = 0;for(int i = 1;i <= n+1;i ++){int l = 0,r = cnt;while(l < r){int mid = (l + r + 1)>>1;if(top[mid] <= b[i]){l = mid;}else r = mid - 1;}if(l == cnt) cnt ++;f[i] = l + 1;top[l + 1] = b[i];ed[f[i]].pb(i);}ed[0].pb(0);memset(g,20,sizeof g);g[0] = 0;for(int i = 1;i <= n+1;i ++){for(int j = 0;j < ed[f[i] - 1].size();j ++){int from = ed[f[i] - 1][j];if(from > i || b[from] > b[i]) continue;pre[from] = sub[i - 1] = 0;for(int k = from + 1;k <= i - 1;k ++){pre[k] = pre[k - 1] + abs(b[k] - b[from]);}for(int k = i - 2;k >= from;k --){sub[k] = sub[k + 1] + abs(b[k + 1] - b[i]);}for(int k = from;k <= i - 1;k ++){g[i] = min(g[i],g[from] + pre[k] + sub[k]);}} }cout<<n - cnt+1<<endl<<g[n + 1]<<endl;}int main() {ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);solve();return 0;
}
