CodeForces 575F Bulbo

洛谷传送门

CF 传送门

提供一个傻逼 \(O(n^2)\) 做法。

首先考虑暴力 dp，设第 \(i\) 轮后在 \(j\) 坐标上的最小花费为 \(f_{i, j}\)，有：

\[f_{i, j} = \min f_{i, k} + |j - k| + \begin{cases} l_i - j & j < l_i \\ 0 & l_i \le j \le r_i \\ j - r_i & j > r_i \end{cases} \]

然后你直观感受一下，除了初始位置，到一个区间的端点肯定不劣。离散化一下，拆绝对值后前后缀 \(\min\) 优化，就是 \(O(n^2)\) 了。

code

// Problem: F. Bulbo
// Contest: Codeforces - Bubble Cup 8 - Finals [Online Mirror]
// URL: https://codeforces.com/problemset/problem/575/F
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 10050;

ll n, m, a[maxn], b[maxn], lsh[maxn], tot, f[2][maxn], g[maxn], h[maxn];

void solve() {
	scanf("%lld%lld", &n, &m);
	lsh[++tot] = m;
	for (int i = 1; i <= n; ++i) {
		scanf("%lld%lld", &a[i], &b[i]);
		lsh[++tot] = a[i];
		lsh[++tot] = b[i];
	}
	sort(lsh + 1, lsh + tot + 1);
	tot = unique(lsh + 1, lsh + tot + 1) - lsh - 1;
	m = lower_bound(lsh + 1, lsh + tot + 1, m) - lsh;
	for (int i = 1; i <= n; ++i) {
		a[i] = lower_bound(lsh + 1, lsh + tot + 1, a[i]) - lsh;
		b[i] = lower_bound(lsh + 1, lsh + tot + 1, b[i]) - lsh;
	}
	mems(f, 0x3f);
	f[0][m] = 0;
	for (int i = 1, o = 1; i <= n; ++i, o ^= 1) {
		mems(f[o], 0x3f);
		g[0] = h[tot + 1] = 1e18;
		for (int j = 1; j <= tot; ++j) {
			g[j] = min(g[j - 1], f[o ^ 1][j] - lsh[j]);
		}
		for (int j = tot; j; --j) {
			h[j] = min(h[j + 1], f[o ^ 1][j] + lsh[j]);
		}
		for (int j = 1; j <= tot; ++j) {
			f[o][j] = min(g[j] + lsh[j], h[j] - lsh[j]) + (j < a[i] ? lsh[a[i]] - lsh[j] : (j > b[i] ? lsh[j] - lsh[b[i]] : 0));
		}
	}
	ll ans = 1e18;
	for (int i = 1; i <= tot; ++i) {
		ans = min(ans, f[n & 1][i]);
	}
	printf("%lld\n", ans);
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}