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BZOJ 1911: [Apio2010]特别行动队 [斜率优化DP]

1911: [Apio2010]特别行动队

Time Limit: 4 Sec  Memory Limit: 64 MB
Submit: 4142  Solved: 1964
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Description

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Input

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Output

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Sample Input

4
-1 10 -20
2 2 3 4

Sample Output

9

HINT

技术分享


f[i]=max{f[j]+...}
随便一化就好了
(a*(s[k]*s[k]-s[j]*s[j])+f[k]-f[j]+b*(s[j]-s[k])) / (2*a*(s[k]-s[j]))
最后是s[i]>=slope(j,k)时k优
s[]是单调的,用单调队列维护这个下凸壳
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;const int N=1e6+5,INF=1e9;typedef long long ll;inline int read(){    char c=getchar();int x=0,f=1;    while(c<0||c>9){if(c==-)f=-1; c=getchar();}    while(c>=0&&c<=9){x=x*10+c-0; c=getchar();}    return x*f;}int n,a,b,c;ll s[N],f[N];inline double slope(int j,int k){    return  (double)(a*(s[k]*s[k]-s[j]*s[j])+f[k]-f[j]+b*(s[j]-s[k]))/(double)(2*a*(s[k]-s[j]));}int q[N],head,tail;void dp(){    head=tail=1;    for(int i=1;i<=n;i++){        while(head<tail&&slope(q[head],q[head+1])<=s[i]) head++;        int j=q[head];        f[i]=f[j]+a*(s[i]-s[j])*(s[i]-s[j])+b*(s[i]-s[j])+c;//printf("f %lld %d\n",f[i],j);        while(head<tail&&slope(q[tail-1],q[tail])>slope(q[tail],i)) tail--;        q[++tail]=i;    }    printf("%lld",f[n]);}int main(){    //freopen("in.txt","r",stdin);    n=read();a=read();b=read();c=read();    for(int i=1;i<=n;i++) s[i]=s[i-1]+read();    dp();}

 

BZOJ 1911: [Apio2010]特别行动队 [斜率优化DP]