http://www.lydsy.com/JudgeOnline/problem.php?id=2244
第$i$个导弹看成一个三元组$(i,h_i,v_i)$
其实就是最长上升子序列的问题。
我们分别求以第$i$个导弹为结尾的最长上升子序列的长度和个数，以及以第$i$个导弹为开头的最长上升子序列的长度和个数。
下面以求以第$i$个导弹为结尾的最长上升子序列的长度和个数为例。
记以第$i$个导弹结尾的最长上升子序列长度为$f[i]$，则：
$$f[i]=Max\{f[j]|j<i,h[j]\geq h[i],v[j]\geq v[i]\}+1$$
所以第1问就是三维偏序，用CDQ分治解决。
第一维由分治的时候左边小于右边保证；
第二维由排序保证；
第三维由数据结构保证；
看程序应该会比较好理解。
记以第$i$个导弹结尾的最长上升子序列长度为$g[i]$，则：
$$g[i]=\sum g[j](f[j]+1=f[i],j<i,h[j]\geq h[i],v[j]\geq v[i])$$
我们可以先按f排序，然后分层DP，还是用CDQ分治。
#include<cstdio> #include<cstdlib> #include<iostream> #include<fstream> #include<algorithm> #include<cstring> #include<string> #include<cmath> #include<queue> #include<stack> #include<map> #include<utility> #include<set> #include<bitset> #include<vector> #include<functional> #include<deque> #include<cctype> #include<climits> #include<complex> //#include<bits/stdc++.h>适用于CF,UOJ，但不适用于pojusing namespace std;typedef long long LL; typedef double DB; typedef pair<int,int> PII; typedef complex<DB> CP;#define mmst(a,v) memset(a,v,sizeof(a)) #define mmcy(a,b) memcpy(a,b,sizeof(a)) #define fill(a,l,r,v) fill(a+l,a+r+1,v) #define re(i,a,b) for(i=(a);i<=(b);i++) #define red(i,a,b) for(i=(a);i>=(b);i--) #define ire(i,x) for(typedef(x.begin()) i=x.begin();i!=x.end();i++) #define fi first #define se second #define m_p(a,b) make_pair(a,b) #define p_b(a) push_back(a) #define SF scanf #define PF printf #define two(k) (1<<(k))template<class T>inline T sqr(T x){return x*x;} template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;} template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}inline int sgn(DB x){if(abs(x)<1e-9)return 0;return(x>0)?1:-1;} const DB Pi=acos(-1.0);int gint(){int res=0;bool neg=0;char z;for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());if(z==EOF)return 0;if(z=='-'){neg=1;z=getchar();}for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());return (neg)?-res:res; } LL gll(){LL res=0;bool neg=0;char z;for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());if(z==EOF)return 0;if(z=='-'){neg=1;z=getchar();}for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());return (neg)?-res:res; }const int maxn=50000;int n,ny,nz; struct Ta{int x,y,z,f,flag;DB g;}a[maxn+100]; int bak[maxn+100];int f1[maxn+100],f2[maxn+100]; DB g1[maxn+100],g2[maxn+100]; int ans;bool cmpx(Ta a,Ta b){return a.x<b.x;} bool cmpyx(Ta a,Ta b){return a.y!=b.y?a.y<b.y:a.x<b.x;}//注意y相同时，x小的在前面！！！ bool cmpf(Ta a,Ta b){return a.f<b.f;}#define lowbit(a) ((a)&(-a)) namespace TREE1{int tree[maxn+100];void update(int a,int v){for(;a<=n;a+=lowbit(a))upmax(tree[a],v);}int ask(int a){int res=0;for(;a>=1;a-=lowbit(a))upmax(res,tree[a]);return res;}void clear(int a){for(;a<=n;a+=lowbit(a))tree[a]=0;}} void CDQf(int l,int r){if(l==r){upmax(a[l].f,1);return;}int i,mid=(l+r)/2;CDQf(l,mid);sort(a+l,a+r+1,cmpyx);re(i,l,r)if(a[i].x<=mid)TREE1::update(a[i].z,a[i].f);else upmax(a[i].f,TREE1::ask(a[i].z)+1);re(i,l,r)if(a[i].x<=mid)TREE1::clear(a[i].z);sort(a+l,a+r+1,cmpx);CDQf(mid+1,r);}namespace TREE2{DB tree[maxn+100];void update(int a,DB v){for(;a<=n;a+=lowbit(a))tree[a]+=v;}DB ask(int a){DB res=0;for(;a>=1;a-=lowbit(a))res+=tree[a];return res;}void clear(int a){for(;a<=n;a+=lowbit(a))tree[a]=0.0;}} void CDQg(int l,int r){if(l==r)return;int i,mid=(l+r)/2;CDQg(l,mid);sort(a+l,a+r+1,cmpyx);re(i,l,r){if(a[i].x<=mid && a[i].flag) TREE2::update(a[i].z,a[i].g);if(a[i].x>mid && !a[i].flag) a[i].g+=TREE2::ask(a[i].z);}re(i,l,r)if(a[i].x<=mid && a[i].flag) TREE2::clear(a[i].z);sort(a+l,a+r+1,cmpx);CDQg(mid+1,r);}int tmpx[maxn+100]; void solve(){int i,j,l1,r1,l2,r2;re(i,1,n)a[i].f=0,a[i].g=0.0;CDQf(1,n);sort(a+1,a+n+1,cmpf);for(l1=r1=1;r1<n && a[r1+1].f==a[l1].f;r1++);re(i,l1,r1)a[i].g=1;for(;r1<n;l1=l2,r1=r2){for(l2=r2=r1+1;r2<n && a[r2+1].f==a[l2].f;r2++);re(i,l1,r1)a[i].flag=1;sort(a+l1,a+r2+1,cmpx);re(i,l1,r2)tmpx[i]=a[i].x,a[i].x=i;CDQg(l1,r2);re(i,l1,r2)a[i].x=tmpx[i];sort(a+l1,a+r2+1,cmpf);re(i,l1,r1)a[i].flag=0;}sort(a+1,a+n+1,cmpx);}int ans1; DB ans2;int main(){freopen("bzoj2244.in","r",stdin);freopen("bzoj2244.out","w",stdout);int i;n=gint();re(i,1,n)a[i].x=i,a[i].y=gint(),a[i].z=gint();ny=0;re(i,1,n)bak[++ny]=a[i].y;sort(bak+1,bak+ny+1);ny=unique(bak+1,bak+ny+1)-bak-1;re(i,1,n)a[i].y=ny-(lower_bound(bak+1,bak+ny+1,a[i].y)-bak)+1;nz=0;re(i,1,n)bak[++nz]=a[i].z;sort(bak+1,bak+nz+1);nz=unique(bak+1,bak+nz+1)-bak-1;re(i,1,n)a[i].z=nz-(lower_bound(bak+1,bak+nz+1,a[i].z)-bak)+1;solve();re(i,1,n)f1[i]=a[i].f,g1[i]=a[i].g;re(i,1,n)a[i].x=n-a[i].x+1,a[i].y=ny-a[i].y+1,a[i].z=nz-a[i].z+1;re(i,1,n/2)swap(a[i],a[n-i+1]);solve();re(i,1,n)a[i].x=n-a[i].x+1,a[i].y=ny-a[i].y+1,a[i].z=nz-a[i].z+1;re(i,1,n/2)swap(a[i],a[n-i+1]);re(i,1,n)f2[i]=a[i].f,g2[i]=a[i].g;re(i,1,n)upmax(ans1,f1[i]);re(i,1,n)if(f1[i]==ans1)ans2+=g1[i];PF("%d\n",ans1);re(i,1,n)if(f1[i]+f2[i]-1!=ans1 || sgn(ans2)==0)PF("0.00000 ");else PF("%0.5lf ",g1[i]*g2[i]/ans2);return 0;}