HDU-4742 Pinball Game 3D 三维LIS

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4742

　　题意：求3维的LIS。。

　　用分治算法搞得，参考了cxlove的题解。。

　　首先按照x排序，然后每个三元组一个编号1-n。接下来只要考虑y和z的值，假设[l,mid]区间已经求好，那么我们对[l,r]区间按照y排序，更新[mid+1,r]区间的最优值时，只要考虑z值了，用树状数组维护z的最长LIS，遇到[mid+1,r]区间的就更新。

 //STATUS:C++_AC_1750MS_5348KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 struct Node{
     int x,y,z,id;
     bool operator <(const Node& a)const{
         return x!=a.x?x<a.x:(y!=a.y?y<a.y:z<a.z);
     }
 }a[N],b[N];
 pii f[N],bit[N];
 int z[N];
 int T,n,m;

 #define lowbit(x) (x&-x)

 inline void update(pii& a,pii& b)
 {
     if(b.first>a.first)a=b;
     else if(b.first==a.first){
         a.second+=b.second;
     }
 }

 inline void add(int x,pii& b)
 {
     for(;x<=m;x+=lowbit(x)){
         update(bit[x],b);
     }
 }

 inline pii query(int x)
 {
     pii ret=make_pair(,);
     for(;x>;x-=lowbit(x)){
         update(ret,bit[x]);
     }
     return ret;
 }

 inline void clear(int x)
 {
     for(;x<=m;x+=lowbit(x)){
         bit[x]=make_pair(,);
     }
 }

 void solve(int l,int r)
 {
     if(l==r)return;
     int i,j,k,mid,cnt=;
     mid=(l+r)>>;
     solve(l,mid);
     for(i=l;i<=r;i++){
         b[cnt]=a[i];
         b[cnt++].x=;
     }
     sort(b,b+cnt);
     for(i=;i<cnt;i++){
         if(b[i].id<=mid){
             add(b[i].z,f[b[i].id]);
         }
         else {
             pii t=query(b[i].z);
             t.first++;
             update(f[b[i].id],t);
         }
     }
     for(i=;i<cnt;i++){
         if(b[i].id<=mid)
             clear(b[i].z);
     }
     solve(mid+,r);
 }

 int main()
 {
   //  freopen("in.txt","r",stdin);
     int i,j;
     pii ans;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         for(i=;i<n;i++){
             scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z);
             z[i]=a[i].z;
         }

         sort(a,a+n);
         sort(z,z+n);
         m=unique(z,z+n)-z;
         for(i=;i<n;i++){
             f[i]=make_pair(,);
             a[i].id=i;
             a[i].z=lower_bound(z,z+m,a[i].z)-z+;
         }
         solve(,n-);
         ans=make_pair(,);
         for(i=;i<n;i++){
             update(ans,f[i]);
         }

         printf("%d %d\n",ans.first,ans.second);
     }
     return ;
 }