BZOJ1010玩具装箱 - 斜率优化dp

传送门

题目分析：

设\(f[i]\)表示装前i个玩具的花费。

列出转移方程：$$f[i] = max{f[j] + ((i - (j + 1)) + sum[i] - sum[j] - L))^2}$$

令\(x[i] = sum[i] + i\), \(P = L + 1\)，上式化为：

\[f[i] = max\{f[j] + (x[i] - x[j] - P)^2\}
\]

列式并化为斜率形式：\(S(i, j) = \frac{(f[i] + x[i]^2 + 2x[i]P) - (f[j] + x[j]^2 + 2x[j]P)}{2(x[i] - x[j])}\)

然后就是斜率dp了。

code

#include<bits/stdc++.h>
using namespace std;
const int N = 50050;
typedef long long ll;
int n;
ll L, c[N];
typedef long long ll;
ll f[N], x[N], sum[N];
int que[N];

inline ll calc(int i, int j){
	return f[j] + (x[i] - x[j] - L) * (x[i] - x[j] - L);
}

inline bool slopeCheck(int i, int j, int k){
	return ((f[i] + x[i] * x[i] + 2 * x[i] * L) - (f[j] + x[j] * x[j] + 2 * x[j] * L)) * 2 * (x[j] - x[k]) <=
		   ((f[j] + x[j] * x[j] + 2 * x[j] * L) - (f[k] + x[k] * x[k] + 2 * x[k] * L)) * 2 * (x[i] - x[j]);
}

int main(){
	freopen("h.in", "r", stdin);
	scanf("%lld%lld", &n, &L);
	for(int i = 1; i <= n; i++) scanf("%lld", &c[i]), sum[i] = sum[i - 1] + c[i], x[i] = sum[i] + i;
	L += 1;
	int head, tail;
	que[head = tail = 1] = 0;
	for(int i = 1; i <= n; i++){
		while(head + 1 <= tail && calc(i, que[head]) >= calc(i, que[head + 1])) head++;
		f[i] = calc(i, que[head]);
		while(head <= tail - 1 && slopeCheck(i, que[tail], que[tail - 1])) tail--;
		que[++tail] = i;
	}
	printf("%lld", f[n]);
	return 0;
}