BZOJ 2244 [SDOI2011]拦截导弹 ——CDQ分治

Posted 2020-09-06 SfailSth ——Salt Fish

三维偏序，直接CDQ硬上。

正反两次CDQ统计结尾的方案数，最后统计即可。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=i;--i)
#define ll long double
#define maxn 50005
struct Mis{
    int id,h,v;
    void print()
    {
        printf("The ID %d High %d Speed %d\n",id,h,v);
    }
}a[maxn];
int f[maxn][2],n,h[maxn],v[maxn],htop,vtop;
ll g[maxn][2];
bool cmp1(Mis a,Mis b){return a.id<b.id;}
bool cmp2(Mis a,Mis b){return a.h<b.h;}
int mx[maxn]; ll cnt[maxn];
void add(int x,int f,ll g)
{
    for (;x<maxn;x+=x&(-x))
        if (mx[x]==f) cnt[x]+=g;
        else if (mx[x]<f) mx[x]=f,cnt[x]=g;
        else break;
}
int qmx(int x)
{
    int ret=0;
    for (;x;x-=x&(-x)) ret=max(ret,mx[x]);
    return ret;
}
ll qcnt(int x,int m)
{
    ll ret=0;
    for (;x;x-=x&(-x))
        if (mx[x]==m) ret+=cnt[x];
    return ret;
}
void del(int x)
{
    for (;x<maxn;x+=x&(-x)) cnt[x]=mx[x]=0;
}
void CDQ(int l,int r,int flag)
{
    if (l==r)
    {
        if (f[a[l].id][flag]<1)
        {
            g[a[l].id][flag]=f[a[l].id][flag]=1;
        }
        return ;
    }
    int mid=l+r>>1;
    sort(a+l,a+r+1,cmp1);
    CDQ(l,mid,flag);
    sort(a+l,a+mid+1,cmp2);
    sort(a+mid+1,a+r+1,cmp2);
    int pl=l;
    F(i,mid+1,r)
    {
        while (a[pl].h<=a[i].h&&pl<=mid) add(a[pl].v,f[a[pl].id][flag],g[a[pl].id][flag]),pl++;
        int tmp=qmx(a[i].v)+1;
        if (tmp==1) continue;
        if (tmp>f[a[i].id][flag])
        {
            f[a[i].id][flag]=tmp;
            g[a[i].id][flag]=qcnt(a[i].v,tmp-1);
        }
        else if (tmp==f[a[i].id][flag])
            g[a[i].id][flag]+=qcnt(a[i].v,tmp-1);
    }
    F(i,l,pl-1) del(a[i].v);
    CDQ(mid+1,r,flag);
}
void print()
{
    int ans=0;
    F(i,1,n) ans=max(ans,f[i][0]);
    printf("%d\n",ans);
    ll sum=0;
    F(i,1,n) if (f[i][0]==ans) sum+=g[i][0]*g[n-i+1][1];
    F(i,1,n)
    {
        if (f[i][0]+f[n-i+1][1]-1==ans) printf("%.5Lf ",g[i][0]*g[n-i+1][1]/sum);
        else printf("0 ");
    }
}
int main()
{
    scanf("%d",&n);htop=vtop=n;
    F(i,1,n)
    {
        scanf("%d%d",&a[i].h,&a[i].v);
        a[i].id=i;h[i]=a[i].h;v[i]=a[i].v;
    }
    sort(h+1,h+htop+1); sort(v+1,v+vtop+1);
    htop=unique(h+1,h+htop+1)-h-1; vtop=unique(v+1,v+vtop+1)-v-1;
    F(i,1,n)
    {
        a[i].h=lower_bound(h+1,h+htop+1,a[i].h)-h;
        a[i].v=lower_bound(v+1,v+vtop+1,a[i].v)-v;
    }
    F(i,1,n)
    {
        a[i].h=htop-a[i].h+1;
        a[i].v=vtop-a[i].v+1; 
    }
    CDQ(1,n,0);
    F(i,1,n)
    {
        a[i].id=n-a[i].id+1;
        a[i].v=vtop-a[i].v+1;
        a[i].h=htop-a[i].h+1;
    }
    CDQ(1,n,1);
    print();
}