Jam's problem again
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 640 Accepted Submission(s): 210
Problem Description
Jam like to solve the problem which on the 3D-axis,given N(1≤N≤100000) points (x,y,z)(1≤x,y,z≤100000)
If two point such as(xi,yi,zi) and (xj,yj,zj) xi≥xj yi≥yj zi≥zj ,
the bigger one level add 1
Ask for the each level of the point.
If two point such as
Ask for the each level of the point.
Input
The first line is T(1≤T≤15) means T Case
For each case
The first line isN means
the number of Point and next there are N line,
each line has (x,y,z)
For each case
The first line is
Output
Output with N line,each line has one number means the lever of point
Sample Input
1 4 10 4 7 10 6 6 8 2 5 7 3 10
Sample Output
1 1 0 0三维偏序，即要求一个三维的点比这个点的小的点有多少个。如果是一维的我们可以怎么做呢？很简单排个序就可以了。如果是二维的呢，同时要求x1<x2和y1<y2点(x1,y1)比点(x2,y2)小。我们可以想到二维树状数组可以很高效率的解决这个问题，但是数据要是有10万，就开不来二维树状数组了。那么可以降维来做。一维排序，二维用树状数组或者线段树来解决，也可以用CDQ分治来做。例如用树状数组，那么按照x排序之后，在树状数组里插入y,边插入边询问，就可以n*logn的效率得到答案。三维的情况呢，同理，第三维可以继续用树状数组或者线段树，但是用树状数组就相当于一个二维树状数组显然会超内存。所以我们可以用动态线段树，不会超内存。如果了用了CDQ分治，那么可以这样一维排序，二维CDQ分治，三维树状数组也可以一维排序，二维CDQ分治，三维CDQ分治一维排序，二维树状数组，三维线段树#include <iostream> #include <string.h> #include <algorithm> #include <math.h> #include <stdio.h> #include <stdlib.h> using namespace std; typedef long long int LL; const int maxn=1e5; struct Node { int x,y,z; int id; int num; }a[maxn+5]; int n; int rt[maxn+5]; int ls[maxn*50]; int rs[maxn*50]; int sum[maxn*50]; int p; int ans1,ans2; int newnode() { ls[p]=rs[p]=sum[p]=0; return p++; } int lowbit(int x) { return x&(-x); } int cmp(Node a,Node b) { if(a.x!=b.x) return a.x<b.x; else { if(a.y!=b.y) return a.y<b.y; else return a.z<b.z; } } int cmp2(Node a,Node b) { return a.id<b.id; } void insert(int &node,int l,int r,int tag) { if(!node) node=newnode(); if(l==r) { sum[node]++; return; } sum[node]++; int mid=(l+r)>>1; if(tag<=mid) insert(ls[node],l,mid,tag); else insert(rs[node],mid+1,r,tag); } int query(int &node,int l,int r,int tag) { if(!node) return 0; if(r<=tag) { return sum[node]; } int mid=(l+r)>>1; if(tag<=mid) return query(ls[node],l,mid,tag); else return sum[ls[node]]+query(rs[node],mid+1,r,tag); } int find(int y,int z) { int res=0; for(int i=y;i>=1;i-=lowbit(i)) { res+=query(rt[i],1,ans2,z); } return res; } void update(int y,int z) { for(int i=y;i<=maxn;i+=lowbit(i)) { insert(rt[i],1,ans2,z); } } void init(int y,int z) { memset(rt,0,sizeof(rt)); for(int i=1;i<=y;i++) { rt[i]=i; ls[i]=rs[i]=sum[i]=0; } p=y+1; } int main() { int t; scanf("%d",&t); while(t--) { scanf("%d",&n); ans1=0; ans2=0; for(int i=1;i<=n;i++) { scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z); ans1=max(ans1,a[i].y); ans2=max(ans2,a[i].z); a[i].id=i;a[i].num=0; } sort(a+1,a+n+1,cmp); for(int i=n-1;i>=1;i--) { if(a[i].x==a[i+1].x&&a[i].y==a[i+1].y&&a[i].z==a[i+1].z) a[i].num=a[i+1].num+1; } init(ans1,ans2); for(int i=1;i<=n;i++) { a[i].num+=find(a[i].y,a[i].z); update(a[i].y,a[i].z); } sort(a+1,a+n+1,cmp2); for(int i=1;i<=n;i++) { printf("%d ",a[i].num); } } return 0; }一维排序，二维CDQ分治，三维树状数组include <iostream> #include <stdlib.h> #include <math.h> #include <algorithm> #include <stdio.h> #include <string.h> using namespace std; typedef long long int LL; const int maxn=1e5; int n; struct Node { int x,y,z; int id; int num; }a[maxn+5],b[maxn+5]; int c[maxn+5]; int cmp(Node a,Node b) { if(a.x==b.x&&a.y==b.y) return a.z<b.z; if(a.x==b.x) return a.y<b.y; return a.x<=b.x; } int cmp2(Node a,Node b) { return a.id<b.id; } int lowbit(int x) { return x&(-x); } void insert(int x,int y) { for(int i=x;i<=maxn;i+=lowbit(i)) c[i]+=y; } int sum(int x) { int s=0; for(int i=x;i>=1;i-=lowbit(i)) s+=c[i]; return s; } void fun(int l,int r) { if(l==r) return; int mid=(l+r)>>1; int j=l; int k=mid+1; fun(l,mid); fun(mid+1,r); for(int i=l;i<=r;i++) { if((a[j].y<=a[k].y||k>r)&&j<=mid) { b[i]=a[j++]; insert(b[i].z,1); } else { b[i]=a[k++]; b[i].num+=sum(b[i].z); } } for(int i=l;i<=r;i++) { if(i<=mid) insert(a[i].z,-1); a[i]=b[i]; } } int main() { int t; scanf("%d",&t); while(t--) { memset(c,0,sizeof(c)); scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z); a[i].id=i;a[i].num=0; } sort(a+1,a+n+1,cmp); for(int i=n-1;i>=1;i--) { if(a[i].x==a[i+1].x&&a[i].y==a[i+1].y&&a[i].z==a[i+1].z) a[i].num=a[i+1].num+1; } fun(1,n); sort(a+1,a+n+1,cmp2); for(int i=1;i<=n;i++) { printf("%d ",a[i].num); } } return 0; }
一维排序，二维CDQ分治，三维CDQ分治#include <iostream> #include <string.h> #include <algorithm> #include <math.h> #include <stdio.h> #include <stdlib.h> using namespace std; typedef long long int LL; const int maxn=1e5; struct Node { int x,y,z; int id,num; int tag; }a[maxn+5],b[maxn+5],c[maxn+5]; int ans[maxn+5]; int n; int cmp(Node a,Node b) { if(a.x==b.x&&a.y==b.y) return a.z<b.z; if(a.x==b.x) return a.y<b.y; return a.x<b.x; } int cmp2(Node a,Node b) { return a.id<b.id; } void fun2(int l,int r) { if(l==r) return; int mid=(l+r)>>1; fun2(l,mid); fun2(mid+1,r); int j=l; int k=mid+1; int res=0; for(int i=l;i<=r;i++) { if((b[j].z<=b[k].z||k>r)&&j<=mid) { c[i]=b[j++]; if(!c[i].tag) res++; } else { c[i]=b[k++]; if(c[i].tag) c[i].num+=res,ans[c[i].id]+=res; } } for(int i=l;i<=r;i++) b[i]=c[i]; } void fun1(int l,int r) { if(l==r) return; int mid=(l+r)>>1; fun1(l,mid); fun1(mid+1,r); int j=l; int k=mid+1; for(int i=l;i<=r;i++) { if((a[j].y<=a[k].y||k>r)&&j<=mid) { b[i]=a[j++]; b[i].tag=0; } else { b[i]=a[k++]; b[i].tag=1; } } for(int i=l;i<=r;i++) a[i]=b[i]; fun2(l,r); } int main() { int t; scanf("%d",&t); while(t--) { scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z); a[i].num=0; a[i].tag=0; a[i].id=i; } memset(ans,0,sizeof(ans)); sort(a+1,a+n+1,cmp); for(int i=n-1;i>=1;i--) if(a[i].x==a[i+1].x&&a[i].y==a[i+1].y&&a[i].z==a[i+1].z) a[i].num=a[i+1].num+1,ans[a[i].id]=ans[a[i+1].id]+1; fun1(1,n); sort(a+1,a+n+1,cmp2); for(int i=1;i<=n;i++) printf("%d ",ans[i]); } return 0; }