hdu_4742_Pinball Game 3D(cdq分治+树状数组)

题目链接：hdu_4742_Pinball Game 3D

题意：

给你n个点，让你求三维的LIS，并且求出有多少种组合能达到LIS。

题解：

求三维的LIS，典型的三维偏序问题，x排序,解决一维，cdq分治y，解决一维，树状数组维护z，解决一维。

注意，cdq中sort不要调用太多，不然会被卡常。

#include<bits/stdc++.h>
#define F(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
typedef pair<int,int>P;

const int N=1e5+7,mod=1<<30;
int t,n,hsh_ed,hsh[N];
P sum[N],dp[N];

struct node
{
    int x,y,z,id;
    bool operator <(const node & b)const{return x<b.x||(x==b.x&&y<b.y)||(x==b.x&&y==b.y&&z<b.z);}
}a[N],b[N];

inline void up(P &a,P b)
{
    if(a.first==b.first)a.second+=b.second;
    else if(a.first<b.first)a=b;
}
inline void add(int x,P c){while(x<=hsh_ed)up(sum[x],c),x+=x&-x;}
inline P ask(int x){P an=P(0,0);while(x)up(an,sum[x]),x-=x&-x;return an;}
inline void back(int x){while(x<=hsh_ed)sum[x]=P(0,0),x+=x&-x;}

void cdq(int l,int r)
{
    if(l==r)return;
    int mid=l+r>>1;
    cdq(l,mid);
    F(i,l,r)b[i]=a[i],b[i].x=0;
    sort(b+l,b+1+r);
    F(i,l,r)
    {
        if(b[i].id<=mid)add(b[i].z,dp[b[i].id]);
        else
        {
            P now=ask(b[i].z);
            if(now.first+1>dp[b[i].id].first)dp[b[i].id]=P(now.first+1,now.second);
            else if(now.first+1==dp[b[i].id].first)dp[b[i].id].second+=now.second;
        }
    }        
    F(i,l,r)if(b[i].id<=mid)back(b[i].z);
    cdq(mid+1,r);
}

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n),hsh_ed=0;
        F(i,1,n)
        {
            scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z);
            hsh[i]=a[i].z;
        }
        sort(a+1,a+1+n);
        sort(hsh+1,hsh+1+n),hsh_ed=unique(hsh+1,hsh+1+n)-hsh;
        F(i,1,n)
        {
            dp[i]=P(1,1),a[i].id=i;
            a[i].z=lower_bound(hsh+1,hsh+1+hsh_ed,a[i].z)-hsh;
        }
        cdq(1,n);
        P ans=P(0,0);
        F(i,1,n)up(ans,dp[i]);
        printf("%d %d\n",ans.first,ans.second%mod);
    }
    return 0;
}