[洛谷P3195][HNOI2008]玩具装箱TOY
题目大意:有n个物体,大小为$c_i$。把第i个到第j个放到一起,容器的长度为$x=j-i+\sum\limits_{k-i}^{j} c_k$,若长度为x,费用为$(x-L)^2$。费用最小.
题解:
$$令:a_i=\sum\limits_{i=1}^{i} c_i$$
$$dp_i=min(dp_j+(a_i+i-a_j-j-L-1)^2)$$
$$(以下称两点斜率为 slope(A,B) )$$
$$令:b_j=a_i+i,d_i=b_i+i+L+1$$
$$\therefore dp_i=dp_j+(b_i-d_j)^2$$
$$展开得:2a_i \cdot b_j+dp_i-a_i^2=dp_j+b_j^2$$
$$令:x_i=2b_i,y_i=dp_i+2b_i^2$$
斜率优化
卡点:无
C++ Code:
#include<cstdio> using namespace std; long long c[50010],f[50010],n,l; int q[50010],h,t,tmp; long long pw(long long i){return i*i;} long long getb(int i){return c[i]+i;} long long getd(int i){return getb(i)-l-1;} long long getx(int i){return getb(i)*2;} long long gety(int i){return f[i]+pw(getb(i));} double slope(int a,int b){ return double(gety(a)-gety(b))/double(getx(a)-getx(b)); } int main(){ scanf("%lld%lld",&n,&l); for (int i=1;i<=n;i++)scanf("%lld",&c[i]),c[i]+=c[i-1]; for (int i=1;i<=n;i++){ while (h<t&&slope(q[h],q[h+1])<=getd(i))h++; tmp=q[h]; f[i]=f[tmp]+pw(getd(i)-getb(tmp)); while (h<t&&slope(q[t-1],q[t])>=slope(q[t],i))t--; q[++t]=i; } printf("%lld\n",f[n]); return 0; }