1010: [HNOI2008]玩具装箱toy
题目：传送门
题解：
很明显的一题动态规划...
f[i]表示1~i的最小花费
那么方程也是显而易见的：f[i]=min(f[j]+(sum[i]-sum[j]+i-(j+1)-L)^2) (j<i)
但是这样的方程写下来要n^2....50000*50000...GG
所以我们要用斜率优化这个神器！！！
设s[i]=sum[i]+i,L+=1;
那么就开始推吧：
f[i]=min(f[i],f[j]+(sum[i]-sum[j]+i-j-1-L)^2)
f[i]=min(f[i],f[j]+(s[i]-s[j]-L)^2)
f[j2]+(s[i]-s[j2]-L)^2<=f[j1]+(s[i]-s[j1]-L)^2
f[j2]+(s[i]-L)^2-2*s[j2]*(s[i]-L)+s[j2]^2<=f[j1]+(s[i]-L)^2-2*s[j1]*(s[i]-L)+s[j1]^2
f[j2]+s[j2]^2-2*s[j2]*(s[i]-L)<=f[j1]+s[j1]^2-2*s[j1]*(s[i]-L)
f[j2]-f[j1]+s[j2]^2-s[j1]^2<=2*s[j2]*(s[i]-L)-2*s[j1]*(s[i]-L)
f[j2]-f[j1]+s[j2]^2-s[j1]^2<=2*s[j2]*(s[i]-L)-2*s[j1]*(s[i]-L)
(f[j2]-f[j1]+s[j2]^2-s[j1]^2)/(s[j2]-s[j1])<=2*(s[i]-L)
代码：
1 #include<cstdio> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cmath> 5 #include<algorithm> 6 using namespace std; 7 typedef long long LL; 8 LL L,sum[51000],s[51000],f[51000],list[51000]; 9 int n; 10 /* 11 f[i]=min(f[i],f[j]+(sum[i]-sum[j]+i-j-1-L)^2) 12 f[i]=min(f[i],f[j]+(s[i]-s[j]-L)^2) 13 f[j2]+(s[i]-s[j2]-L)^2<=f[j1]+(s[i]-s[j1]-L)^2 14 f[j2]+(s[i]-L)^2-2*s[j2]*(s[i]-L)+s[j2]^2<=f[j1]+(s[i]-L)^2-2*s[j1]*(s[i]-L)+s[j1]^2 15 f[j2]+s[j2]^2-2*s[j2]*(s[i]-L)<=f[j1]+s[j1]^2-2*s[j1]*(s[i]-L) 16 f[j2]-f[j1]+s[j2]^2-s[j1]^2<=2*s[j2]*(s[i]-L)-2*s[j1]*(s[i]-L) 17 f[j2]-f[j1]+s[j2]^2-s[j1]^2<=2*s[j2]*(s[i]-L)-2*s[j1]*(s[i]-L) 18 (f[j2]-f[j1]+s[j2]^2-s[j1]^2)/(s[j2]-s[j1])<=2*(s[i]-L) 19 */ 20 double slope(int j1,int j2) 21 { 22 return (f[j2]-f[j1]+s[j2]*s[j2]-s[j1]*s[j1])/(s[j2]-s[j1]); 23 } 24 int main() 25 { 26 memset(f,0,sizeof(f)); 27 scanf("%d%lld",&n,&L);sum[0]=0; 28 LL x;for(int i=1;i<=n;i++){scanf("%lld",&x);sum[i]=sum[i-1]+x;} 29 for(int i=1;i<=n;i++)s[i]=sum[i]+i; 30 L++; 31 int head=1,tail=1;list[1]=0; 32 for(int i=1;i<=n;i++) 33 { 34 while(head<tail && slope(list[head],list[head+1])<=2.0*double(s[i]-L))head++; 35 int j=list[head]; 36 f[i]=f[j]+(s[i]-s[j]-L)*(s[i]-s[j]-L); 37 while(head<tail && slope(list[tail],i)<slope(list[tail-1],list[tail]))tail--; 38 list[++tail]=i; 39 } 40 printf("%lld ",f[n]); 41 return 0; 42 } 43