题解 亿个社区

传送门

考虑暴力是 \(O(n^2k)\) 的
考虑暴力这份码：

for (int i=1; i<=n; ++i)
	for (int j=1; j<=k; ++j)
		for (int l=0; l<i; ++l)
			f[i][j]=min(f[i][j], f[l][j-1]+sqr(t[i]-t[l+1]));

发现最后这个转移可以把括号拆开，然后动态凸包优化
复杂度 \(O(nk\log n)\)

点击查看代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f3f3f3f3f
#define N 100010
#define fir first
#define sec second
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, k;
ll t[N];
inline ll sqr(ll t) {return t*t;}

namespace force{
	ll f[10100][110];
	void solve() {
		memset(f, 0x3f, sizeof(f));
		f[0][0]=0;
		for (int i=1; i<=n; ++i)
			for (int j=1; j<=k; ++j)
				for (int l=0; l<i; ++l)
					f[i][j]=min(f[i][j], f[l][j-1]+sqr(t[i]-t[l+1]));
		printf("%lld\n", f[n][k]);
	}
}

namespace task1{
	ll f[10010][110];
	struct vec{
		mutable ll k, b; mutable double p;
		vec(ll x, ll y):k(x),b(y),p(0){}
	};
	inline bool operator < (vec a, vec b) {return a.k>b.k;}
	inline bool operator < (vec a, double b) {return a.p<b;}
	struct convex:multiset<vec, less<>>{
		const double inf=1e30;
		inline double qmeet(vec a, vec b) {return double(b.b-a.b)/double(a.k-b.k);}
		bool isect(iterator a, iterator b) {
			if (b==end()) return a->p=inf, 0;
			if (a->k==b->k) a->p = a->b > b->b ? -inf : inf;
			else a->p = qmeet(*a, *b);
			return a->p >= b->p;
		}
		void ins(vec t) {
			auto z=insert(t), y=z++, x=y;
			while (isect(y, z)) z=erase(z);
			if (x!=begin() && isect(--x, y)) isect(x, y=erase(y));
			while ((y=x)!=begin() && (--x)->p >= y->p) isect(x, y=erase(y));
		}
		ll qmin(ll x) {
			if (empty()) return INF;
			auto it=lower_bound(x);
			return it->k*x+it->b;
		}
	}con[110], tem;
	// set<pair<ll, ll>> s[110];
	void solve() {
		con[0].ins({-2*t[1], sqr(t[1])});
		// s[0].insert({-2*t[1], sqr(t[1])});
		for (int i=1; i<=n; ++i) {
			for (int j=1; j<=min(i, k); ++j) f[i][j]=sqr(t[i])+con[j-1].qmin(t[i]); //, cout<<con[j-1].qmin(t[i])<<endl;
			for (int j=1; j<=min(i, k); ++j) con[j].ins({-2*t[i+1], sqr(t[i+1])+f[i][j]});
			// for (int j=1; j<=min(i, k); ++j) {
			// 	f[i][j]=INF;
			// 	for (auto it:s[j-1]) f[i][j]=min(f[i][j], sqr(t[i])+t[i]*it.fir+it.sec);
			// }
			// for (int j=1; j<=min(i, k); ++j) s[j].insert({-2*t[i+1], sqr(t[i+1])+f[i][j]});
		}
		printf("%lld\n", f[n][k]);
	}
}

signed main()
{
	freopen("community.in", "r", stdin);
	freopen("community.out", "w", stdout);

	n=read(); k=read();
	for (int i=1; i<=n; ++i) t[i]=read();
	// force::solve();
	task1::solve();

	return 0;
}