1911: [Apio2010]特别行动队
Time Limit: 4 Sec Memory Limit: 64 MBSubmit: 4048 Solved: 1913
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Description
Input
Output
Sample Input
4
-1 10 -20
2 2 3 4
-1 10 -20
2 2 3 4
Sample Output
9
【题解】
首先很容易想到用前缀和，下面的sum表示前缀和。
然后写出状态转移方程：f[i]=max{f[j]+a(sum[i]-sum[j])^2+b(sum[i]-sum[j])+c}
假设j比k更优，得到斜率表达式(f[j]+a*sum[j]^2-b*sum[j])-(f[k]+a*sum[k]^2-b*sum[k])/(sum[j]-sum[k])>2a*sum[i]
然后斜率优化走起。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<ctime> 6 #include<cmath> 7 #include<algorithm> 8 using namespace std; 9 typedef long long ll; 10 #define MAXN 1000100 11 #define FILE "read" 12 #define up(i,j,n) for(ll i=j;i<=n;i++) 13 ll n,a,b,c,l,r,x[MAXN],sum[MAXN],q[MAXN],f[MAXN]; 14 namespace INIT{ 15 char buf[1<<15],*fs,*ft; 16 inline char getc() {return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;} 17 inline ll read() { 18 ll x=0,f=1; char ch=getc(); 19 while(!isdigit(ch)) {if(ch=='-') f=-1; ch=getc();} 20 while(isdigit(ch)) {x=x*10+ch-'0'; ch=getc();} 21 return x*f; 22 } 23 }using namespace INIT; 24 inline double slop(ll j,ll k) {return (double)((f[j]+a*sum[j]*sum[j]-b*sum[j])-(f[k]+a*sum[k]*sum[k]-b*sum[k]))/(double)(sum[j]-sum[k]);} 25 int main(){ 26 freopen(FILE".in","r",stdin); 27 freopen(FILE".out","w",stdout); 28 n=read(); a=read(); b=read(); c=read(); 29 up(i,1,n) x[i]=read(),sum[i]=sum[i-1]+x[i]; 30 up(i,1,n){ 31 while(l<r&&slop(q[l],q[l+1])>2*a*sum[i]) l++; 32 ll t=q[l]; 33 f[i]=f[t]+a*(sum[i]-sum[t])*(sum[i]-sum[t])+b*(sum[i]-sum[t])+c; 34 while(l<r&&slop(q[r],i)>slop(q[r-1],q[r])) r--; 35 q[++r]=i; 36 } 37 printf("%lld ",f[n]); 38 return 0; 39 }