dp+线段树+单调栈：ABC480G

G - Zonal Score Minimization

image-20261017214138631

设 fif_i 表示前 ii 个数的满足题意的答案

显然有：

fi=min⁡j≤i−2(fj+min⁡k∈[j+1,i]ak−max⁡k∈[j+1,i]ak)f_i=\min_{j\le i -2}(f_j+\min_{k\in[j+1,i]}a_k-\max_{k\in[j+1,i]}a_k)

我们考虑拿数据结构来维护。后缀min、max之类的我们可以直接拿单调栈来求，然后拿个线段树来维护区间加和区间最小值即可。

时间复杂度 O(nlog⁡n)O(n\log n)

需要注意的是要把 f1=+∞f_1=+\infty

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#include<bits/stdc++.h>
#include<cassert>
using namespace std;bool Mbe;
#ifdef LOCAL
 #define debug(...) fprintf(stdout, ##__VA_ARGS__)
#else
 #define debug(...) void(0)
#endif
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
#define Z(x) (x)*(x)
#define pb push_back
#define fi first
#define se second
//#define M
//#define mo
#define N 200010
int n, m, i, j, k, T;
int a[N], f[N], ans, lmn[N], lmx[N], rt; 
stack<int>zmn, zmx; 

struct Segment_tree {
	int mn[N << 2], ls[N << 2], rs[N << 2], tag[N << 2], tot; 
	void build(int &x, int l, int r) {
		if(!x) x = ++tot; 
		if(l == r) return ; 
		int mid = (l + r) >> 1; 
		build(ls[x], l, mid); 
		build(rs[x], mid + 1, r); 
	}
	void push_down(int k) {
		mn[ls[k]] += tag[k]; tag[ls[k]] += tag[k]; 
		mn[rs[k]] += tag[k]; tag[rs[k]] += tag[k]; 
		assert(mn[k] == min(mn[ls[k]], mn[rs[k]])); 
		tag[k] = 0; 
	}
	void push_up(int k) {
		assert(tag[k] == 0); 
		mn[k] = min(mn[ls[k]], mn[rs[k]]); 
	}
	void add(int k, int l, int r, int x, int y, int z) {
		if(k == 1) debug("## [%lld %lld] += %lld\n", x, y, z); 
		if(l >= x && r <= y) {
			mn[k] += z; tag[k] += z; 
			return ; 
		}
		int mid = (l + r) >> 1; 
		push_down(k); 
		if(x <= mid) add(ls[k], l, mid, x, y, z); 
		if(y >= mid + 1) add(rs[k], mid + 1, r, x, y, z); 
		push_up(k); 
	}
	int qry(int k, int l, int r, int x, int y) {
		if(l >= x && r <= y) return mn[k]; 
		push_down(k); 
		int mid = (l + r) >> 1, ans = 1e18; 
		if(x <= mid) ans = min(ans, qry(ls[k], l, mid, x, y)); 
		if(y >= mid + 1) ans = min(ans, qry(rs[k], mid + 1, r, x, y)); 
		push_up(k); 
		return ans; 
	}
	void print(int k, int l, int r) {
		if(l == r) return printf("%lld ", mn[k]), void(); 
		int mid = (l + r) >> 1; 
		push_down(k); 
		print(ls[k], l, mid); 
		print(rs[k], mid + 1, r); 
		push_up(k); 
	}
}Seg; 

bool Med; signed main() {
	#ifdef LOCAL
	  freopen("in.txt", "r", stdin);
	  freopen("out.txt", "w", stdout);
	  atexit([]{fprintf(stderr,"%.lfms\n%lfMB\n",clock()*1000./CLOCKS_PER_SEC,(&Mbe-&Med)/1024./1024);});
	#endif
//	srand(time(NULL));
//	T = read();
//	while(T--) {
//
//	}
	n = read(); 
	Seg.build(rt, 0, n); 
	Seg.add(1, 0, n, 1, 1, 1e18); 
	for(i = 1; i <= n; ++i) {
		a[i] = read();  
		while(!zmn.empty() && a[i] < a[zmn.top()]) {
			j = zmn.top(); zmn.pop(); 
//			debug("[%lld %lld] %lld\n", lmn[j] - 1, j - 1, -(-a[j])); 
			Seg.add(1, 0, n, lmn[j] - 1, j - 1, -(-a[j])); 
		}
		j = (zmn.empty() ? 1 : zmn.top() + 1); 
		lmn[i] = j; zmn.push(i); 
		debug("min[%lld %lld] -> %lld\n", j, i, a[i]); 
		Seg.add(1, 0, n, j - 1, i - 1, -a[i]); 
		while(!zmx.empty() && a[i] > a[zmx.top()]) {
			j = zmx.top(); zmx.pop(); 
//			debug("[%lld %lld] %lld\n", lmx[j] - 1, j - 1, -(-a[j])); 
			Seg.add(1, 0, n, lmx[j] - 1, j - 1, -(a[j])); 
		}
		j = (zmx.empty() ? 1 : zmx.top() + 1); 
		debug("max[%lld %lld] -> %lld\n", j, i, a[i]); 
		Seg.add(1, 0, n, j - 1, i - 1, a[i]); 
		lmx[i] = j; zmx.push(i); 
		if(i == 1) continue; 
//		Seg.print(1, 0, n); debug("\n"); 
		f[i] = Seg.qry(1, 0, n, 0, i - 2); 
		debug("f[%lld] = %lld\n", i, f[i]); 
		Seg.add(1, 0, n, i, i, f[i]); 
	}
	printf("%lld", f[n]); 
	return 0;
}