hdu 5126 stars cdq分治
stars
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 282 Accepted Submission(s): 68
Problem Description
John loves to see the sky. A day has Q times. Each time John will find a new star in the sky, or he wants to know how many stars between
(x1,y1,z1) and
(x2,y2,z2) .
Input
The first line contains a single integer
T(1≤T≤10) (the data for
Q>100 less than 6 cases),, indicating the number of test cases.
The first line contains an integer Q(1≤Q≤50000) ,indicating how many times in a day.
Next Q lines contain some integers, first input an integer A(1≤A≤2) .If A=1 then input 3 integers x, y and z, indicating a coordinate of one star.. If A=2 then input 6 integers x1,y1,z1,x2,y2,z2(1≤x,y,z,x1,y1,z1,x2,y2,z2≤109,x1≤x2,y1≤y2,z1≤z2) .
The first line contains an integer Q(1≤Q≤50000) ,indicating how many times in a day.
Next Q lines contain some integers, first input an integer A(1≤A≤2) .If A=1 then input 3 integers x, y and z, indicating a coordinate of one star.. If A=2 then input 6 integers x1,y1,z1,x2,y2,z2(1≤x,y,z,x1,y1,z1,x2,y2,z2≤109,x1≤x2,y1≤y2,z1≤z2) .
Output
For each “A=2”,output an integer means how many stars in such a section.
Sample Input
2 11 1 1 1 1 2 1 1 1 1 1 1 1 2 2 2 1 1 1 2 2 1 1 1 2 2 2 1 3 3 3 1 4 4 4 1 5 5 5 1 6 6 6 2 1 1 1 6 6 6 2 3 3 3 6 6 6 11 1 1 1 1 2 1 1 1 1 1 1 1 2 2 2 1 1 1 2 2 1 1 1 2 2 2 1 3 3 3 1 4 4 4 1 5 5 5 1 6 6 6 2 1 1 1 6 6 6 2 3 3 3 6 6 6
Sample Output
1 3 7 4 1 3 7 4
用kd树做超时了。网上看到是cdq分治的，看了下ppt。
http://wenku.baidu.com/link?url=VwuZ4ij8GABr5FOVcuU2yyMelwPSa_sXahA9lgVu0bQTrF2J3GuZGgHXE7LinI0-EulsxJJXZ3oMFL4dlIvOZdDlrIudCptcpawthnAMlj3
还看了下别人的解法
最后我的做法大致是这样的，但是比较慢
把一个询问拆成8个查询（容斥咯），用树状数组维护坐标z从0到需要查询位置点的个数，当然z要先离散化。
然后就是用cdq分治，使得每次查询都是有效点了。
思想就是划分一个个集合，使得每个集合里插入点在已经被处理的维度上，坐标都小于询问的坐标值
1. 因为插入和查询的点是有顺序的，所以其实应该看成是四维的。第一次应该先按操作的次序排序（其实就是不用排序了）做cdq分治
2.在cdq分治的第一次里，分治使得点按x轴排序，因为按左右两个部分，可以知道左边部分的次序<右边部分的操作次序，那么把左边的插入和右边的查询放到一起，那么新的集合里在次序这个维度上，查询的就高于插入的了。又因为分治按x排序，那么就可以线性的把新的集合按照x轴大小为第一关键字，次序为第二关键字的方式排序
3.在新的集合里，对y轴进行排序，因为左边的插入操作的点x<右边操作的x，那么满足左边插入的点的y小于右边当前查询的点y，那么在更新树状数组中左边插入点z处值，然后更新查询即可，由于树状数组每次要清空，所以插入的点都需要删除
对于树状数组的清空操作有两种方式：
1. 时间戳标记，用T[]，tag，如果T[]=tag说明这个位置的值是新的，否则置成0，让T[]=tag，每次要清空的时候tag++即可
2.删除插入的点
#include<cstdio>
#include<algorithm>
using namespace std;
struct Node{
int x,y,z,p,t;
Node(){}
Node(int x,int y,int z,int p,int t):x(x),y(y),z(z),p(p),t(t){}
};
#define maxn 450007
Node star[maxn];
Node star2[maxn];
Node star3[maxn];
int tree[maxn];
int lowbit(int i){
return i&(-i);
}
void add(int i,int x){
while(i<maxn){
tree[i]+=x;
i+=lowbit(i);
}
}
int get(int i){
int ans = 0;
while(i>0){
ans += tree[i];
i -= lowbit(i);
}
return ans;
}
int res[maxn];
void CDQ2(int l,int r){
if(l == r) return ;
int mid = (l+r)/2;
CDQ2(l,mid);
CDQ2(mid+1,r);
int l1 =l, r1 = mid+1;
while(r1 <= r){
while(l1 <= mid && star2[l1].y <= star2[r1].y){
if(star2[l1].t == 0) add(star2[l1].z,1);
l1++;
}
if(star2[r1].t != 0){
res[star2[r1].p] += get(star2[r1].z)*star2[r1].t;
}
r1++;
}
while(l1 > l){
--l1;
if(star2[l1].t == 0) add(star2[l1].z,-1);
}
l1 = l, r1 = mid+1;
for(int i = l;i <= r; i++){
if((l1 <= mid && star2[l1].y <= star2[r1].y) || r1 > r )
star3[i] = star2[l1++];
else star3[i] = star2[r1++];
}
for(int i = l; i <= r; i++)
star2[i] = star3[i];
}
void CDQ1(int l,int r){
if(l == r) return ;
int mid = (l+r)/2;
CDQ1(l,mid);
CDQ1(mid+1,r);
int l1 = l, r1 = mid+1,n = 0;
while(r1 <= r){
while(star[l1].t != 0 && l1 <= mid) l1++;
while(star[r1].t == 0 && r1 <= r) r1++;
if(r1 > r) break;
if((star[l1].x <= star[r1].x && l1 <= mid)|| r1 > r)
star2[n++] = star[l1++];
else star2[n++] = star[r1++];
}
if(n > 0) CDQ2(0,n-1);
l1 = l, r1 = mid+1;
for(int i = l;i <= r; i++){
if((star[l1].x <= star[r1].x && l1 <= mid)|| r1 > r)
star3[i] = star[l1++];
else star3[i] = star[r1++];
}
for(int i = l;i <= r; i++)
star[i] = star3[i];
}
int compn(Node a,Node b){
return a.p < b.p;
}
int compz(Node a,Node b){
return a.z < b.z;
}
int main(){
int t,q,a,x1,y1,z1,x2,y2,z2;
scanf("%d",&t);
while(t--){
int n = 0;
scanf("%d",&q);
while(q--){
scanf("%d",&a);
if(a == 1){
scanf("%d%d%d",&x1,&y1,&z1);
star[n++] = Node(x1,y1,z1,n,0);
}
else {
scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
star[n++] = Node(x2,y2,z2,n,1);
star[n++] = Node(x2,y1-1,z2,n,-1);
star[n++] = Node(x1-1,y2,z2,n,-1);
star[n++] = Node(x2,y2,z1-1,n,-1);
star[n++] = Node(x1-1,y1-1,z2,n,1);
star[n++] = Node(x1-1,y2,z1-1,n,1);
star[n++] = Node(x2,y1-1,z1-1,n,1);
star[n++] = Node(x1-1,y1-1,z1-1,n,-1);
}
}
for(int i = 0;i < n; i++)
res[i] = 0;
sort(star,star+n,compz);
a = 1;
star[n].z = -1;
for(int i = 0;i < n; i++){
if(star[i].z != star[i+1].z)
star[i].z = a++;
else star[i].z = a;
}
sort(star,star+n,compn);
CDQ1(0,n-1);
sort(star,star+n,compn);
for(int i = 0;i < n; i++){
if(star[i].t != 0){
int ans = 0;
for(int j = 0;j < 8; j++)
ans += res[i+j];
printf("%d\n",ans);
i += 7;
}
}
}
return 0;
}