Slope Trick 总结

Slope Trick 总结

注意:Slope Trick 并不是斜率优化,斜率优化的英文是 Convex Hull Trick

算法适用性

Slope Trick 通常用于维护具有如下性质的函数:

  • 连续。
  • 是分段一次函数。
  • 是凸函数。
  • 每一段的斜率较小(通常为 \(O(n)\)),且均为整数。

常常用于优化动态规划。

不失一般性,约定本文中的函数均为下凸,且称以上所有性质为性质 \(P\)

注意到,两个满足性质 \(P\) 的函数相加,得到的函数依然满足性质 \(P\)

算法思想

维护在某个 \(x_0\) 处的 \(f(x_0),k_0\) 以及函数每个斜率变化点的集合。具体地,如果函数在 \(x\) 位置斜率增加了 \(\Delta k\),就在集合中插入 \(\Delta k\)\(x\)

性质 \(P\) 十分优美,它使得我们可以快速维护很多操作:

  • 相加:将 \(f(x_0),k_0\) 直接相加,斜率变化点的集合直接合并。常用于加一次函数、绝对值函数。
  • 取前缀/后缀 \(\min\):去掉 \(k<0\)\(k>0\) 的部分。
  • \(\min,\operatorname{argmin}\):提取 \(k=0\) 部分。
  • 平移:维护 \(f(x_0),k_0\) 变化,斜率变化点在全局打平移标记。
  • 翻转:维护 \(f(x_0),k_0\) 变化,斜率变化点在全局打翻转标记。

例题:P4597 序列 sequence

\(f_{i,j}\) 表示考虑 \([1,i]\) 的前缀,把最后一个数变为 \(j\) 的最小操作次数。显然有转移方程:

\[f_{i,j}=\min_{k\le j}\{f_{i-1,k}\}+|a_i-j| \]

\(F_i(j)=f_{i,j}\)。显然,\(F_i\) 可以由 \(F_{i-1}\) 先取前缀 \(\min\) 再加绝对值函数得到。所求即为 \(\min F_n\)

由于 \(F_i\) 满足性质 \(P\),使用堆维护斜率变化点即可。

代码
//By: Luogu@rui_er(122461)
#include <bits/stdc++.h>
#define rep(x,y,z) for(ll x=y;x<=z;x++)
#define per(x,y,z) for(ll x=y;x>=z;x--)
#define debug printf("Running %s on line %d...\n",__FUNCTION__,__LINE__)
#define fileIO(s) do{freopen(s".in","r",stdin);freopen(s".out","w",stdout);}while(false)
using namespace std;
typedef long long ll;
const ll N = 5e5+5; 

ll n, a[N], ans;
priority_queue<ll> q;
template<typename T> void chkmin(T& x, T y) {if(x > y) x = y;}
template<typename T> void chkmax(T& x, T y) {if(x < y) x = y;}

int main() {
	scanf("%lld", &n);
	rep(i, 1, n) {
		scanf("%lld", &a[i]);
		if(!q.empty() && q.top() > a[i]) {
			ans += q.top() - a[i];
			q.pop();
			q.push(a[i]);
		}
		q.push(a[i]);
	}
	printf("%lld\n", ans);
	return 0;
}

四倍经验:CF713C Sonya and Problem Wihtout a LegendCF13C SequenceP2893 [USACO08FEB] Making the Grade G
与该题几乎完全一致。

习题:LCP24 数字游戏

Slope Trick 简单题,也有对顶堆等做法。

习题:P3642 [APIO2016] 烟火表演

可并堆维护 Slope Trick。

代码
// Problem: P3642 [APIO2016] 烟火表演
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3642
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

//By: OIer rui_er
#include <bits/stdc++.h>
#define rep(x, y, z) for(int x = (y); x <= (z); ++x)
#define per(x, y, z) for(int x = (y); x >= (z); --x)
#define debug(format...) fprintf(stderr, format)
#define fileIO(s) do {freopen(s".in", "r", stdin); freopen(s".out", "w", stdout);} while(false)
#define endl '\n'
using namespace std;
typedef long long ll;

mt19937 rnd(std::chrono::duration_cast<std::chrono::nanoseconds>(std::chrono::system_clock::now().time_since_epoch()).count());
int randint(int L, int R) {
    uniform_int_distribution<int> dist(L, R);
    return dist(rnd);
}

template<typename T> void chkmin(T& x, T y) {if(x > y) x = y;}
template<typename T> void chkmax(T& x, T y) {if(x < y) x = y;}

template<int mod>
inline unsigned int down(unsigned int x) {
	return x >= mod ? x - mod : x;
}

template<int mod>
struct Modint {
	unsigned int x;
	Modint() = default;
	Modint(unsigned int x) : x(x) {}
	friend istream& operator>>(istream& in, Modint& a) {return in >> a.x;}
	friend ostream& operator<<(ostream& out, Modint a) {return out << a.x;}
	friend Modint operator+(Modint a, Modint b) {return down<mod>(a.x + b.x);}
	friend Modint operator-(Modint a, Modint b) {return down<mod>(a.x - b.x + mod);}
	friend Modint operator*(Modint a, Modint b) {return 1ULL * a.x * b.x % mod;}
	friend Modint operator/(Modint a, Modint b) {return a * ~b;}
	friend Modint operator^(Modint a, int b) {Modint ans = 1; for(; b; b >>= 1, a *= a) if(b & 1) ans *= a; return ans;}
	friend Modint operator~(Modint a) {return a ^ (mod - 2);}
	friend Modint operator-(Modint a) {return down<mod>(mod - a.x);}
	friend Modint& operator+=(Modint& a, Modint b) {return a = a + b;}
	friend Modint& operator-=(Modint& a, Modint b) {return a = a - b;}
	friend Modint& operator*=(Modint& a, Modint b) {return a = a * b;}
	friend Modint& operator/=(Modint& a, Modint b) {return a = a / b;}
	friend Modint& operator^=(Modint& a, int b) {return a = a ^ b;}
	friend Modint& operator++(Modint& a) {return a += 1;}
	friend Modint operator++(Modint& a, int) {Modint x = a; a += 1; return x;}
	friend Modint& operator--(Modint& a) {return a -= 1;}
	friend Modint operator--(Modint& a, int) {Modint x = a; a -= 1; return x;}
	friend bool operator==(Modint a, Modint b) {return a.x == b.x;}
	friend bool operator!=(Modint a, Modint b) {return !(a == b);}
};

const ll N = 6e5 + 5;

ll n, m, f[N], d[N], deg[N], ans;

struct LeftistTreeNode {
    ll lc, rc, val, dis;
    LeftistTreeNode(ll val = 0, ll dis = 0) : lc(0), rc(0), val(val), dis(dis) {}
};
struct LeftistTree {
    LeftistTreeNode t[N];
    ll rt[N], sz;
    ll newnode(ll val) {
        t[++sz] = LeftistTreeNode(val);
        return sz;
    }
    ll merge(ll u, ll v) {
        if(!u || !v) return u | v;
        if(t[u].val < t[v].val) swap(u, v);
        t[u].rc = merge(t[u].rc, v);
        if(t[t[u].lc].dis < t[t[u].rc].dis) swap(t[u].lc, t[u].rc);
        t[u].dis = t[t[u].rc].dis + 1;
        return u;
    }
    ll merge(ll u, ll v, ll args...) {
        return merge(merge(u, v), args);
    }
    ll seek(ll u) {
        return t[u].val;
    }
    void pop(ll& u) {
        u = merge(t[u].lc, t[u].rc);
    }
}leftree;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cin >> n >> m;
    rep(i, 2, n + m) cin >> f[i] >> d[i];
    rep(i, 2, n + m) ++deg[f[i]], ans += d[i];
    per(u, n + m, 2) {
        while(deg[u]-- > 1) leftree.pop(leftree.rt[u]);
        ll r = leftree.seek(leftree.rt[u]);
        leftree.pop(leftree.rt[u]);
        ll l = leftree.seek(leftree.rt[u]);
        leftree.pop(leftree.rt[u]);
        leftree.rt[u] = leftree.merge(leftree.rt[u], leftree.newnode(l + d[u]), leftree.newnode(r + d[u]));
        leftree.rt[f[u]] = leftree.merge(leftree.rt[f[u]], leftree.rt[u]);
    }
    while(deg[1]--) leftree.pop(leftree.rt[1]);
    while(leftree.rt[1]) {
        ans -= leftree.seek(leftree.rt[1]);
        leftree.pop(leftree.rt[1]);
    }
    cout << ans << endl;
    return 0;
}
posted @ 2024-02-16 14:09  rui_er  阅读(1615)  评论(0编辑  收藏  举报