bzoj 2244: [SDOI2011]拦截导弹

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #define M 100009
 using namespace std;
 struct data
 {
     int x,y,z,f[];
     double sum[];
 }a[M],b[M];
 struct ss
 {
     int w;
     double su;
 }shu[M];
 int n,yy[M],zz[M],yl,zl,ans,tot,sta[M];
 double sum1;
 bool cmp(data a1,data a2)
 {
     if(a1.y==a2.y)
       return a1.x<a2.x;
     return a1.y<a2.y;
 }
 bool cmp1(data a1,data a2)
 {
     return a1.x>a2.x;
 }
 void updata(int i,int p)
 {
     int a1=a[i].z;
     for(;a1<=zl;a1+=a1&-a1)
       if(shu[a1].w<a[i].f[p])
         {
             if(shu[a1].w==)
               sta[++tot]=a1;
             shu[a1].w=a[i].f[p];
             shu[a1].su=a[i].sum[p];
         }
       else if(shu[a1].w==a[i].f[p])
              shu[a1].su+=a[i].sum[p];
     return;
 }
 void ask(int i,int p)
 {
     int a1=a[i].z;
     for(;a1;a1-=a1&-a1)
       if(a[i].f[p]<=shu[a1].w&&shu[a1].w)
         {
             a[i].f[p]=shu[a1].w+;
             a[i].sum[p]=shu[a1].su;
         }
       else if(a[i].f[p]==shu[a1].w+)
              a[i].sum[p]+=shu[a1].su;
     return;
 }
 void solve(int l,int r,int p)
 {
     if(l==r)
       {
         if(a[l].f[p]==)
           a[l].f[p]=a[l].sum[p]=;
         return;
       }
     int mid=(l+r)>>;
     int l1=l,l2=mid+;
     for(int i=l;i<=r;i++)
       if(a[i].x<=mid)
         b[l1++]=a[i];
       else
         b[l2++]=a[i];
     for(int i=l;i<=r;i++)
       a[i]=b[i];
     solve(l,mid,p);
     sort(a+l,a+mid+,cmp);
     int st=l;
     for(int i=mid+;i<=r;i++)
       {
         for(;a[st].y<=a[i].y&&st<=mid;st++)
           updata(st,p);
         ask(i,p);
       }
     for(int i=;i<=tot;i++)
       shu[sta[i]].w=shu[sta[i]].su=;
     tot=;
     solve(mid+,r,p);
 }
 int main()
 {
     scanf("%d",&n);
     for(int i=;i<=n;i++)
       {
         a[i].x=i;
         scanf("%d%d",&a[i].y,&a[i].z);
         yy[i]=a[i].y;
         zz[i]=a[i].z;
       }
     sort(yy+,yy+n+);
     sort(zz+,zz+n+);
     yl=unique(yy+,yy+n+)-yy-;
     zl=unique(zz+,zz+n+)-zz-;
     for(int i=;i<=n;i++)
       {
         a[i].y=yl-(lower_bound(yy+,yy+yl+,a[i].y)-yy)+;
         a[i].z=zl-(lower_bound(zz+,zz+zl+,a[i].z)-zz)+;
       }
     sort(a+,a+n+,cmp);
     solve(,n,);
     for(int i=;i<=n;i++)
       {
         a[i].x=n-a[i].x+;
         a[i].y=yl-a[i].y+;
         a[i].z=zl-a[i].z+;
       }
     sort(a+,a+n+,cmp);
     solve(,n,);
     sort(a+,a+n+,cmp1);
     for(int i=;i<=n;i++)
       if(a[i].f[]>ans)
         {
             ans=a[i].f[];
             sum1=a[i].sum[];
         }
       else if(a[i].f[]==ans)
              sum1+=a[i].sum[];
     printf("%d\n",ans);
     for(int i=;i<=n;i++)
       if(a[i].f[]+a[i].f[]-==ans)
         printf("%.5lf ",a[i].sum[]*a[i].sum[]/sum1);
       else
         printf("0.00000 ");
     return ;
 }

首先第一问明显是一个三维偏序集，速度，高度，时间，用CDQ分治做，然后我们把它反过来，在做一边CDQ分治，这两遍求出来的方案数组相乘，就是过这个点的方案数。