bzoj 1911: [Apio2010]特别行动队 -- 斜率优化
1911: [Apio2010]特别行动队
Time Limit: 4 Sec Memory Limit: 64 MBDescription
![](http://www.lydsy.com/JudgeOnline/images/1911_1.jpg)
Input
![](http://www.lydsy.com/JudgeOnline/images/1911_2.jpg)
Output
![](http://www.lydsy.com/JudgeOnline/images/1911_3.jpg)
Sample Input
4
-1 10 -20
2 2 3 4
-1 10 -20
2 2 3 4
Sample Output
9
HINT
Source
dp方程:
如果j>k且j比k更优
#include<map> #include<cmath> #include<queue> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define ll long long #define N 1000100 #define db double char xB[1<<15],*xS=xB,*xTT=xB; #define getc() (xS==xTT&&(xTT=(xS=xB)+fread(xB,1,1<<15,stdin),xS==xTT)?0:*xS++) #define isd(c) (c>='0'&&c<='9') inline int read(){ char xchh; int xaa; while(xchh=getc(),!isd(xchh));(xaa=xchh-'0'); while(xchh=getc(),isd(xchh))xaa=xaa*10+xchh-'0';return xaa; } int n,a,b,c,x[N],q[N],l,r,t; ll f[N],sum[N]; inline ll sqr(ll x){return x*x;} inline db cal(int j,int k){return (db)(f[j]+a*sqr(sum[j])-b*sum[j]-f[k]-a*sqr(sum[k])+b*sum[k])/(db)(2*a*(sum[j]-sum[k]));} int main() { scanf("%d%d%d%d",&n,&a,&b,&c); for(int i=1;i<=n;i++) x[i]=read(); for(int i=1;i<=n;i++) sum[i]=sum[i-1]+x[i]; for(int i=1;i<=n;i++) { while(l<r&&cal(q[l],q[l+1])<sum[i]) l++; t=q[l]; f[i]=f[t]+a*sqr(sum[i]-sum[t])+b*(sum[i]-sum[t])+c; while(l<r&&cal(q[r-1],q[r])>cal(q[r],i)) r--; q[++r]=i; } printf("%lld\n",f[n]); return 0; }
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