【斜率DP】BZOJ 1911:特别行动队

1911: [Apio2010]特别行动队

Time Limit: 4 Sec  Memory Limit: 64 MB
Submit: 3006  Solved: 1360
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Description

Input

Output

Sample Input

4
-1 10 -20
2 2 3 4

Sample Output

9

HINT

　　不开long long见祖宗。。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cmath>
 4 #include<cstdio>
 5
 6 #define maxn 1000001
 7
 8 using namespace std;
 9
10 long long sum[maxn],f[maxn];
11
12 int a,b,c,n,que[maxn],head=1,tail=1;
13
14 double K(int x,int y){return (double)(sum[x]+sum[y])/2.0+(double)(f[y]-f[x])/(2.0*a*(sum[y]-sum[x]))-(double)b/(2.0*a);}
15
16 long long F(int x){return a*x*x+b*x+c;}
17
18 void DP()
19 {
20     que[tail]=0;
21     for(int i=1;i<=n;i++)
22     {
23         while(head<tail && sum[i] >= K(que[head],que[head+1]))head++;
24         int sd=que[head];
25         f[i]=f[sd]+F(sum[i]-sum[sd]);
26         while(head<tail && K(que[tail],i) <= K(que[tail-1],que[tail]))tail--;
27         que[++tail]=i;
28     }
29     printf("%lld",f[n]);
30 }
31
32 int main()
33 {
34     scanf("%d%d%d%d",&n,&a,&b,&c);
35     for(int i=1;i<=n;i++)scanf("%lld",&sum[i]),sum[i]+=sum[i-1];
36     DP();
37     return 0;
38 }