斜率dp cdq 分治

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <algorithm>
 5 using namespace std;
 6  
 7 #define maxn 100008
 8 #define LL long long 
 9 
10 long long f[maxn];
11 int a[maxn],b[maxn];
12 int n;
13 bool flag=1;
14 
15 void read(int &x){
16     char ch;
17     for (ch=getchar();ch<‘0‘||ch>‘9‘;ch=getchar()); x=ch-48;
18     for (ch=getchar();ch>=‘0‘&&ch<=‘9‘;ch=getchar()) x=x*10+ch-48;
19 }
20  
21 void init(){
22     read(n);
23     for (int i=1;i<=n;i++) { read(a[i]); read(b[i]); if (b[i]) flag=0; }
24     for (int i=1;i<=n;i++) f[i]=(LL)1<<60;
25 }
26  
27 void force(){
28     for (int i=1;i<=n;i++)
29         for (int j=0;j<=i-1;j++)
30             if (a[j]>=b[i])
31                 f[i]=min(f[i],f[j]+(LL)(a[i]-a[j])*(a[i]-a[j]));
32     
33 }
34  
35  int q[maxn],rk[maxn];
36 bool cmp(int i,int j){
37     return a[i]<a[j] ;
38 }
39 
40 long long kx(int i,int j){
41     return 2*(a[i]-a[j]);
42 }
43 
44 long long ky(int i,int j){
45     long long ans=1LL*a[i]*a[i]+f[i]-1LL*a[j]*a[j]-f[j]; 
46     return ans;
47 }
48 
49 bool cmp1(int i,int j,int k){
50     return ky(k,j)*kx(j,i)<=kx(k,j)*ky(j,i);
51 }
52 
53 bool cmp2(int i,int j,int k){
54     return ky(i,j)>=k*kx(i,j); 
55 }
56 
57  void solve(int l,int r){
58     if (l==r) return; 
59     int mid=(l+r)>>1;
60     solve(l,mid);
61     for (int i=l;i<=r;i++) rk[i]=i;
62     sort(rk+l,rk+r+1,cmp);
63     int h=1,t=0;
64     for (int i=l;i<=r;i++)
65     {
66         if (rk[i]<=mid) {
67             while (h<t&&cmp1(q[t-1],q[t],rk[i])) t--;
68             q[++t]=rk[i];
69         } else {
70             while (h<t&&cmp2(q[h],q[h+1],a[rk[i]])) h++;
71             f[rk[i]]=min(f[rk[i]],f[q[h]]+1LL*(a[rk[i]]-a[q[h]])*(a[rk[i]]-a[q[h]]));
72         }
73     }
74     solve(mid+1,r);
75  }
76 int main(){
77     init();
78     if (n<=1000) force(); 
79     if (flag) solve(0,n);
80     printf("%.4f",sqrt(f[n]));
81 }