[BZOJ2244][SDOI2011]拦截导弹（dp+cdq分治）

题目描述

传送门

题解

这题真tm麻烦

其实dp是很简单的，令f(i)表示以i为结尾的最长非升子序列长度，g(i)表示以i结尾的最长非升子序列的方案数，h(i)和k(i)分别表示以i为开头的

然后对于每一个i，判断f(i)+h(i)-1是否就是最长答案；如果是，那么g(i)*k(i)就是这个点出现的次数

dp的过程用两个cdq做一下就行，实际上就是一个三维偏序，按照时间分治，每次按照横坐标排序，然后纵坐标用bit查询一下

这题好像精度有什么问题…全改成double就过了…

代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define N 100005
#define LL long long

int n;
double ans,down;
int lsh
,LSH;
struct data{int x,y,id;}q
;
double f
,h
,g
,k
;

namespace FG
{
double C
,D
;
double maxf,maxg;
data a
,b
;int ch
;
int cmp(data a,data b)
{
return a.x>b.x;
}
void add(int loc,double valf,double valg)
{
for (int i=loc;i<=LSH;i+=i&-i)
if (valf>C[i]) C[i]=valf,D[i]=valg;
else if (valf==C[i]) D[i]+=valg;
}
void query(int loc)
{
maxf=0,maxg=0;
for (int i=loc;i>=1;i-=i&-i)
if (C[i]>maxf) maxf=C[i],maxg=D[i];
else if (C[i]==maxf) maxg+=D[i];
}
void cover(int loc)
{
for (int i=loc;i<=LSH;i+=i&-i)
C[i]=0,D[i]=0;
}
void cdq(int l,int r)
{
if (l==r)
{
f[l]=max(f[l],1.0);
g[l]=max(g[l],1.0);
return;
}
int mid=(l+r)>>1;
cdq(l,mid);
int acnt=0,bcnt=0;
for (int i=l;i<=mid;++i) a[++acnt]=q[i];
for (int i=mid+1;i<=r;++i) b[++bcnt]=q[i];
sort(a+1,a+acnt+1,cmp);sort(b+1,b+bcnt+1,cmp);
int pa=1,pb=1,tot=0;
while (pb<=bcnt)
{
while (pa<=acnt&&a[pa].x>=b[pb].x)
{
add(LSH-a[pa].y+1,f[a[pa].id],g[a[pa].id]);
ch[++tot]=LSH-a[pa].y+1;
++pa;
}
query(LSH-b[pb].y+1);
int now=b[pb].id;
if (f[now]<maxf+1) f[now]=maxf+1,g[now]=maxg;
else if (f[now]==maxf+1) g[now]+=maxg;
++pb;
}
for (int i=1;i<=tot;++i) cover(ch[i]);
cdq(mid+1,r);
}
void solve()
{
cdq(1,n);
}
}
namespace HK
{
double C
,D
;
double maxh,maxk;
data a
,b
;int ch
;

int cmp(data a,data b)
{
return a.x>b.x;
}
void add(int loc,double valh,double valk)
{
for (int i=loc;i<=LSH;i+=i&-i)
if (valh>C[i]) C[i]=valh,D[i]=valk;
else if (valh==C[i]) D[i]+=valk;
}
void query(int loc)
{
maxh=0,maxk=0;
for (int i=loc;i>=1;i-=i&-i)
if (C[i]>maxh) maxh=C[i],maxk=D[i];
else if (C[i]==maxh) maxk+=D[i];
}
void cover(int loc)
{
for (int i=loc;i<=LSH;i+=i&-i)
C[i]=0,D[i]=0;
}
void cdq(int l,int r)
{
if (l==r)
{
h[l]=max(h[l],1.0);
k[l]=max(k[l],1.0);
return;
}
int mid=(l+r)>>1;
cdq(mid+1,r);
int acnt=0,bcnt=0;
for (int i=l;i<=mid;++i) a[++acnt]=q[i];
for (int i=mid+1;i<=r;++i) b[++bcnt]=q[i];
sort(a+1,a+acnt+1,cmp);sort(b+1,b+bcnt+1,cmp);
int pa=acnt,pb=bcnt,tot=0;
while (pa>=1)
{
while (pb>=1&&b[pb].x<=a[pa].x)
{
add(b[pb].y,h[b[pb].id],k[b[pb].id]);
ch[++tot]=b[pb].y;
--pb;
}
query(a[pa].y);
int now=a[pa].id;
if (h[now]<maxh+1) h[now]=maxh+1,k[now]=maxk;
else if (h[now]==maxh+1) k[now]+=maxk;
--pa;
}
for (int i=1;i<=tot;++i) cover(ch[i]);
cdq(l,mid);
}
void solve()
{
cdq(1,n);
}
}
int main()
{
scanf("%d",&n);
for (int i=1;i<=n;++i)
{
scanf("%d%d",&q[i].x,&q[i].y);
lsh[++LSH]=q[i].y;q[i].id=i;
}
sort(lsh+1,lsh+LSH+1);LSH=unique(lsh+1,lsh+LSH+1)-lsh-1;
for (int i=1;i<=n;++i) q[i].y=lower_bound(lsh+1,lsh+LSH+1,q[i].y)-lsh;
for (int i=1;i<=n;++i)
f[1]=1;g[1]=1;
FG::solve();
for (int i=1;i<=n;++i) ans=max(ans,f[i]);
printf("%.0lf\n",ans);
for (int i=1;i<=n;++i)
if (f[i]==ans) down+=g[i];
h
=1;k
=1;
HK::solve();
double zero=0.0;
for (int i=1;i<=n;++i)
if (f[i]+h[i]-1==ans)
printf("%.5lf%c",g[i]*k[i]/down," \n"[i==n]);
else printf("%.5lf%c",zero," \n"[i==n]);
}