1911: [Apio2010]特别行动队
Time Limit: 4 Sec Memory Limit: 64 MBSubmit: 3969 Solved: 1873
Description
Input
Output
Sample Input
4
-1 10 -20
2 2 3 4
-1 10 -20
2 2 3 4
Sample Output
9
HINT
Source
斜率优化DP
f[分组末尾]=最优解
f[i] = max{f[j]+A(s[i]-s[j])^2+B(s[i]-s[j])+C} (0<=j<i) (s[]是战斗力前缀和)
对于任意两个决策点j>k，假设j比k更优，则有：
f[j]+A(s[i]-s[j])^2+B(s[i]-s[j])+C >= f[k]+A(s[i]-s[k])^2+B(s[i]-s[k])+C
设G(j)=f[j]+A*s[j]^2-B*s[j] ，H(j)=-2*A*s[j]
由于H(j)>H(k)，得到[G(j)-G(k)]/[H(j)-H(k)]>=-s[i]
1 /*by SilverN*/ 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cstdio> 6 #include<cmath> 7 #define LL long long 8 using namespace std; 9 const int mxn=1e6+10; 10 int read(){ 11 int x=0,f=1;char ch=getchar(); 12 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 13 while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();} 14 return x*f; 15 } 16 LL f[mxn]; 17 int n; 18 int a,b,c; 19 int x[mxn]; 20 LL s[mxn]; 21 LL gdp(int i,int x){ 22 return f[x]+a*(s[i]-s[x])*(s[i]-s[x])+b*(s[i]-s[x])+c; 23 } 24 LL gup(int x,int y){ 25 return f[x]-f[y]+a*(s[x]*s[x]-s[y]*s[y])-b*(s[x]-s[y]); 26 } 27 LL gdown(int x,int y){ 28 return -2*a*s[x]+2*a*s[y]; 29 } 30 int hd=0,tl=0; 31 int q[mxn]; 32 int main(){ 33 n=read(); 34 a=read();b=read();c=read(); 35 int i,j; 36 for(i=1;i<=n;i++){ 37 x[i]=read(); 38 s[i]=s[i-1]+x[i]; 39 } 40 for(i=1;i<=n;i++){ 41 while(hd<tl && gup(q[hd],q[hd+1])<=-s[i]*gdown(q[hd],q[hd+1]))hd++; 42 f[i]=gdp(i,q[hd]); 43 while(hd<tl && gup(q[tl],i)*gdown(q[tl-1],q[tl])>=gup(q[tl-1],q[tl])*gdown(q[tl],i))tl--; 44 q[++tl]=i; 45 } 46 printf("%lld ",f[n]); 47 return 0; 48 }