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 + sumlimits_{k = i}^{j}ck )
求最小花费
题解思路
( dp[i] = min(dp[j] + (i - j - 1 + sumlimits_{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
", dp[n]);
}