洛谷 P3195 [HNOI2008]玩具装箱TOY

题意简述

有n个物体，第i个长度为ci

将n个物体分为若干组，每组必须连续

如果把i到j的物品分到一组，则该组长度为 \( j - i + \sum\limits_{k = i}^{j}ck \)

求最小花费

题解思路

\( dp[i] = min(dp[j] + (i - j - 1 + \sum\limits_{k = i}^{j}ck)) \)

然后斜率优化，单调队列维护

代码

#include <cstdio>
using namespace std;
typedef long long ll;
int n, l, h, t;
int q[50010];
ll sum[50010], dp[50010];
ll sqr(ll x) {return x * x; }
int s1(int x) {return sum[x] + x; }
int s2(int x) {return s1(x) + l + 1; }
double calc(int i, int j) {return (double)(sqr(s2(i)) + dp[i] - sqr(s2(j)) - dp[j]) / (s1(i) - s1(j)); }
int main()
{
	scanf("%d%d", &n, &l);
	for (register int i = 1; i <= n; ++i)
	{
		scanf("%d", &sum[i]);
		sum[i] += sum[i - 1];
	}
	h = t = 1;
	for (register int i = 1; i <= n; ++i)
	{
		while (h < t && calc(q[h], q[h + 1]) <= 2 * s1(i)) ++h;
		dp[i] = dp[q[h]] + sqr(i - q[h] - 1 + sum[i] - sum[q[h]] - l);
		while (h < t && calc(i, q[t - 1]) < calc(q[t - 1], q[t])) --t;
		q[++t] = i;
	}
	printf("%lld\n", dp[n]);
}