[HNOI2008]玩具装箱TOY

problem

---

---

Solution

20’DP:

$dp[i]=min(dp[j]+(sum[i]-sum[j-1]+i-j-L{)}^2)$

code:

#include<bits/stdc++.h>
#define N 500005
#define int long long
using namespace std;
int sqr(int x){
	return x * x;
}
int n, L, a[N], sum[N], f[N];
signed main(){
	scanf("%lld%lld", &n, &L);
	for(int i = 1; i <= n; i ++) scanf("%lld", &a[i]), sum[i] = a[i] + sum[i - 1];
	memset(f, 0x3f, sizeof f);
	f[0] = 0;
	for(int i = 1; i <= n; i ++){
		for(int j = 1; j <= i; j ++){
			f[i] = min(f[i], f[j - 1] + sqr(sum[i] - sum[j - 1] + i - j - L));
		}
	}
	printf("%lld", f[n]);
	return 0;
}

斜率优化

可以变成
$dp[i]=min(dp[j]+(sum[i]-sum[j]+i-j-L-1{)}^2)$

令 $a[i]=sum[i]+i,b[i]=sum[i]+i+L+1$

$dp[i]=min(dp[j]+(a[i]-b[j]{)}^2)$

$dp[i]=dp[j]+a[i{]}^2+b[j{]}^2-2\cdot a[i]\cdot b[j]$

$2\cdot a[i]\cdot b[j]+dp[i]-a[i{]}^2=dp[j]+b[j{]}^2$

把$b[j]$看做$x$，$dp[j]+b[j{]}^2$看做$y$

$y=2\cdot a[i]x+dp[i]-a[i{]}^2$

这就是一条直线的解析式，$y=kx+b,k=2\cdot a[i],b=dp[i]-a[i{]}^2$

要找一个点$P(b[j],dp[j]+b[j{]}^2)$使得上面的斜率为2*a[i]的直线过这个点且与$y$轴截距$(b=dp[i]-a[i{]}^2)$最小

因为斜率$k=2\cdot a[i]$是固定的，所以要求的就是最小的$b$,把它加上$a[i{]}^2$就是$dp[i]$的值

---

大概就是这样

就变成了用一条斜率为$k=2\cdot a[i]$的直线从下到上（或从右到左）扫过去，碰到的第一个点后把直线和$y$轴的截距求出来加上$a[i]$就是答案

很明显就是维护一个下凸壳(1/4)

---

然后就单调队列维护嘛。。

设$Slope(i,j)$为 $P_i,P_j$ 的斜率

如果$Slope(Q_{head},Q_{head+1})<=2\cdot a[i]$就 $head++$

$Slope(Q_{tail},Q_{tail-1})>=Slope(Q_{tail-1},i)$就 $tail--$

code:

#include<bits/stdc++.h>
#define N 1000005
#define int long long//hei hei
#define double long double //开long double 防止精度丢失
using namespace std;
int q[N], n, L, sum[N], dp[N];
double a(int i) {return sum[i] + i;}
double b(int i) {return sum[i] + i + L + 1;}
double x(int i) {return b(i);}
double y(int i) {return dp[i] + b(i) * b(i);}
double Slope(int i, int j) {return (y(i) - y(j)) / (x(i) - x(j));}
signed main(){
	scanf("%lld%lld", &n, &L);
	for(int i = 1; i <= n; i ++) scanf("%lld", &sum[i]), sum[i] += sum[i - 1];
	int l = 1, r = 1;
	for(int i = 1; i <= n; i ++){
		while(l < r && Slope(q[l], q[l + 1]) < 2 * a(i)) l ++;
		int j = q[l]; 
		dp[i] = dp[j] + (a(i) - b(j)) * (a(i) - b(j));//截距就是把x=2*b[j]代入斜率为2*a[i]的直线，解除b，加上a[i]就是dp[i],化简完就是这样， 或者直接转移也可以
		while(l < r && Slope(q[r - 1], q[r]) > Slope(i, q[r - 1])) r --;
		q[++ r] = i;
	}
	printf("%lld", dp[n]);
	return 0;
}