hdu 5126 stars cdq分治

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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).

Output

For each “A=2”,output an integer means how many stars in such a section.

Sample Input

2111 1 1 12 1 1 1 1 1 11 2 2 21 1 1 22 1 1 1 2 2 21 3 3 31 4 4 41 5 5 51 6 6 62 1 1 1 6 6 62 3 3 3 6 6 6111 1 1 12 1 1 1 1 1 11 2 2 21 1 1 22 1 1 1 2 2 21 3 3 31 4 4 41 5 5 51 6 6 62 1 1 1 6 6 62 3 3 3 6 6 6

Sample Output

13741374

题目意思：刚开始三维空间没有点，给m个操作，a=1插入一个点，a=2询问一个子空间中有几个点

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 450007Node 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;}

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