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=100010;
const int INF=0x3f3f3f3f;
const int MOD=100000,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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(0,0);
    for(;x>0;x-=lowbit(x)){
        update(ret,bit[x]);
    }
    return ret;
}

inline void clear(int x)
{
    for(;x<=m;x+=lowbit(x)){
        bit[x]=make_pair(0,0);
    }
}

void solve(int l,int r)
{
    if(l==r)return;
    int i,j,k,mid,cnt=0;
    mid=(l+r)>>1;
    solve(l,mid);
    for(i=l;i<=r;i++){
        b[cnt]=a[i];
        b[cnt++].x=0;
    }
    sort(b,b+cnt);
    for(i=0;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=0;i<cnt;i++){
        if(b[i].id<=mid)
            clear(b[i].z);
    }
    solve(mid+1,r);
}

int main()
{
  //  freopen("in.txt","r",stdin);
    int i,j;
    pii ans;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=0;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=0;i<n;i++){
            f[i]=make_pair(1,1);
            a[i].id=i;
            a[i].z=lower_bound(z,z+m,a[i].z)-z+1;
        }
        solve(0,n-1);
        ans=make_pair(0,0);
        for(i=0;i<n;i++){
            update(ans,f[i]);
        }

        printf("%d %d
",ans.first,ans.second);
    }
    return 0;
}