bzoj 1911 [Apio2010]特别行动队(斜率优化+DP)

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思路:

斜率优化。
设f[i]表示将前i个分组的最优值,则有转移方程式:
f[i]=max{ f[j]+a*(s[i]-s[j])^2+b*(s[i]-s[j])+c }
经过化简得到:f[i]=max{ (f[j]+a*s[j]^2-b*s[j])-2*a*s[i]*s[j] } + a*s[i]^2+b*s[i]+c
单调队列维护上凸包即可。
y[j] = (f[j]+a*s[j]^2-b*s[j])
x[j] = s[j]
min p = y[j]-2*a*s[i]*x[j] 因为a是负的 所以斜率为正 是上凸包


now.x = s[j]
now.y = y[i] = (f[i]+a*s[i]^2-b*s[i]) = {(f[j]+a*s[j]^2-b*s[j])-2*a*s[i]*s[j]+ a*s[i]^2+b*s[i]+c} + a*s[i]^2-b*s[i] = {(q[L].y)-2*a*s[i]*q[L].x} + 2*a*s[i]^2+c

答案就是 q[R].y-a*s[n]*s[n]+b*s[n] 

代码一:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 #define mem(a) memset(a,0,sizeof(a))
 5 #define mp(x,y) make_pair(x,y)
 6 const int INF = 0x3f3f3f3f;
 7 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
 8 inline ll read(){
 9     ll x=0,f=1;char ch=getchar();
10     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
11     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
12     return x*f;
13 }
14 //////////////////////////////////////////////////////////////////////////
15 const int maxn = 1e6+10;
16 
17 struct node{
18     ll x,y;
19 }now,q[maxn];
20 
21 int n,x;
22 ll a,b,c,s[maxn];
23 
24 ll cross(node A,node B,node C){
25     return (B.x-A.x)*(C.y-A.y) - (C.x-A.x)*(B.y-A.y);
26 }
27 
28 int main(){
29     n=read();
30     a=read(),b=read(),c=read();
31     s[0] = 0;
32     for(int i=1; i<=n; i++){
33         x=read();
34         s[i] = s[i-1]+x;
35     }
36 
37     int L=0,R=0;
38     for(int i=1; i<=n; i++){
39         while(L<R && q[L+1].y-2*a*s[i]*q[L+1].x >= q[L].y-2*a*s[i]*q[L].x) L++;
40         while(L<R && q[L].y-2*a*s[i]*q[L].x <= q[L+1].y-2*a*s[i]*q[L+1].x) L++;
41         now.x = s[i];
42         now.y = q[L].y-2*a*s[i]*q[L].x+2*a*s[i]*s[i]+c;    
43         while(L<R && cross(q[R-1],now,q[R]) <= 0) R--; // 为什么now跑到中间去嘞?是上凸包
44         q[++R] = now;
45          
46     }
47 
48     cout << q[R].y-a*s[n]*s[n]+b*s[n] << endl;
49 
50     return 0;
51 }

 

代码二:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 #define mem(a) memset(a,0,sizeof(a))
 5 #define mp(x,y) make_pair(x,y)
 6 const int INF = 0x3f3f3f3f;
 7 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
 8 inline ll read(){
 9     ll x=0,f=1;char ch=getchar();
10     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
11     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
12     return x*f;
13 }
14 //////////////////////////////////////////////////////////////////////////
15 const int maxn = 1e6+10;
16 
17 int n,x,q[maxn];
18 ll a,b,c,s[maxn],f[maxn];
19 
20 ll getup(int j,int k){
21     return f[j]-f[k]+a*(s[j]*s[j]-s[k]*s[k])+b*(s[k]-s[j]);
22 }
23 
24 ll getdown(int j,int k){
25     return (s[j]-s[k]);
26 }
27 
28 int main(){
29     n=read();
30     a=read(),b=read(),c=read();
31     s[0] = 0;
32     for(int i=1; i<=n; i++){
33         x=read();
34         s[i] = s[i-1]+x;
35     }
36 
37     int L=0,R=0;
38     for(int i=1; i<=n; i++){
39         while(L<R && getup(q[L+1],q[L]) >= s[i]*2*a*getdown(q[L+1],q[L])) L++; // a是负的
40         int j = q[L];
41         f[i] = f[j] + a*(s[i]-s[j])*(s[i]-s[j]) + b*(s[i]-s[j]) + c;
42         while(L<R && getup(i,q[R])*getdown(q[R],q[R-1]) >= getup(q[R],q[R-1])*getdown(i,q[R])) R--;
43         q[++R] = i;    
44     }
45 
46     cout << f[n] << endl;
47 
48     return 0;
49 }

 

posted @ 2017-02-23 15:01  _yxg123  阅读(121)  评论(0编辑  收藏  举报