【BZOJ1010】【HNOI2008】玩具装箱toy 解题报告
wangyuchen
2018-03-17 12:37:50
## 思路：
这题不难想。
首先我们先推出一个普通的dp方程：
$f_i = min \{ f_j+(i-j-1+sum_i-sum_j-L)^2\}$
然后就推一推式子了：
$f_j+(i-j-1+sum_i-sum_j-L)^2 < f_k+(i-k-1+sum_i-sum_k-L)^2$
令$num_i = i+sum_i$
令$C = L+1$
$f_j+(num_i-num_j-L-1)^2 < f_k+(num_i-num_j-L-1)^2$
$f_j+{num_i}^2-2*num_i*(num_j+C)+(num_j+C)^2$
$<f_k+{num_i}^2-2*num_i*(num_k+C)+(num_k+C)^2$
$f_j+(num_j+C)^2-f_k-(num_k+C)^2 < $
$2*num_i*(num_j-num_k)$
${f_j+(num_j+C)^2-f_k-(num_k+C)^2 \over 2*(num_j-num_k)} < num_i$
接下来就可以用斜率优化dp了。
## 代码：
```cpp
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 50010;
inline long long sqr(long long x) { return x*x; }
long long a[N];
long long sum[N];
long long num[N], C;
long long f[N];
double calc(int j, int k) { return (double)(f[j]+sqr(num[j]+C)-f[k]-sqr(num[k]+C))/(double)(2*(num[j]-num[k])); }
int Q[N], hd, tl;
int main()
{ int n;
long long m;
scanf("%d%lld", &n, &m);
for(int i=1; i<=n; i++)
{ scanf("%lld", &a[i]);
sum[i] = sum[i-1]+a[i];
}
for(int i=1; i<=n; i++) num[i] = sum[i]+i;
C = m+1;
Q[hd = 0] = 0;
tl = 1;
for(int i=1; i<=n; i++)
{ while(hd < tl-1 && calc(Q[hd+1], Q[hd]) <= num[i]) hd++;
f[i] = f[Q[hd]] + sqr(num[i]-num[Q[hd]]-C);
while(hd < tl-1 && calc(i, Q[tl-1]) <= calc(Q[tl-1], Q[tl-2])) tl--;
Q[tl++] = i;
}
printf("%lld\n", f[n]);
return 0;
}
```