HDU 5126 stars KDTree

stars

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 980    Accepted Submission(s): 311

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

 

KD树裸题，最开始想着先不重构KD树，交上去试试，如果超时了再改。结果一直WA，一直在找BUG都找不到，实在没办法就试着重构KD树，于是就每1000个节点重构一次KD树，直接A了，没想到明明超时OJ却判成WA。

#include <bits/stdc++.h>#define maxn 100010#define inf 0x7fffffffusing namespace std;int n, m, root, cmpd;struct data{    int l, r, d[3], mx[3], mn[3], sum, w; // sum:以此为根节点数目和  w:节点上数字    data() {}    data(int a, int b, int c, int val)    {        l = r = 0;        d[0] = mx[0] = mn[0] = a;        d[1] = mx[1] = mn[1] = b;        d[2] = mx[2] = mn[2] = c;        w = sum = val;    }} tr[maxn];void update(int x){    int l = tr[x].l, r = tr[x].r;    if (l) // 维护以此为根的子树各个维度范围与节点数目和    {        tr[x].mx[0] = max(tr[x].mx[0], tr[l].mx[0]);        tr[x].mn[0] = min(tr[x].mn[0], tr[l].mn[0]);        tr[x].mx[1] = max(tr[x].mx[1], tr[l].mx[1]);        tr[x].mn[1] = min(tr[x].mn[1], tr[l].mn[1]);        tr[x].mx[2] = max(tr[x].mx[2], tr[l].mx[2]);        tr[x].mn[2] = min(tr[x].mn[2], tr[l].mn[2]);        tr[x].sum += tr[l].sum;    }    if (r)    {        tr[x].mx[0] = max(tr[x].mx[0], tr[r].mx[0]);        tr[x].mn[0] = min(tr[x].mn[0], tr[r].mn[0]);        tr[x].mx[1] = max(tr[x].mx[1], tr[r].mx[1]);        tr[x].mn[1] = min(tr[x].mn[1], tr[r].mn[1]);        tr[x].mx[2] = max(tr[x].mx[2], tr[r].mx[2]);        tr[x].mn[2] = min(tr[x].mn[2], tr[r].mn[2]);        tr[x].sum += tr[r].sum;    }}int cmp(data a, data b){    if (a.d[cmpd] == b.d[cmpd])    {        if (a.d[(cmpd + 1) % 3] == b.d[(cmpd + 1) % 3])        {            return a.d[(cmpd + 2) % 3] < b.d[(cmpd + 2) % 3];        }        return a.d[(cmpd + 1) % 3] < b.d[(cmpd + 1) % 3];    }    return a.d[cmpd] < b.d[cmpd];}int build(int l, int r, int d){    int mid = (l + r) >> 1;    cmpd = d;    //不对整体排序，只求第n小的元素并放在其位置上    nth_element(tr + l, tr + mid, tr + r + 1, cmp);    tr[mid].mx[0] = tr[mid].mn[0] = tr[mid].d[0];    tr[mid].mx[1] = tr[mid].mn[1] = tr[mid].d[1];    tr[mid].mx[2] = tr[mid].mn[2] = tr[mid].d[2];    tr[mid].sum = tr[mid].w;    tr[mid].l = 0;    tr[mid].r = 0;    if (l < mid)    {        tr[mid].l = build(l, mid - 1, (d + 1) % 3);    }    if (r > mid)    {        tr[mid].r = build(mid + 1, r, (d + 1) % 3);    }    update(mid);    return mid;}void kdinsert(int now){    if (now == root)    {        return;    }    int D, p;    D = 0;    p = root;    while (true)    {        //插入到P中，维护信息        if (tr[now].mx[0] > tr[p].mx[0])        {            tr[p].mx[0] = tr[now].mx[0];        }        if (tr[now].mx[1] > tr[p].mx[1])        {            tr[p].mx[1] = tr[now].mx[1];        }        if (tr[now].mx[2] > tr[p].mx[2])        {            tr[p].mx[2] = tr[now].mx[2];        }        if (tr[now].mn[0] < tr[p].mn[0])        {            tr[p].mn[0] = tr[now].mn[0];        }        if (tr[now].mn[1] < tr[p].mn[1])        {            tr[p].mn[1] = tr[now].mn[1];        }        if (tr[now].mn[2] < tr[p].mn[2])        {            tr[p].mn[2] = tr[now].mn[2];        }        tr[p].sum += tr[now].sum;        if (tr[p].d[0] == tr[now].d[0] && tr[p].d[1] == tr[now].d[1] && tr[p].d[2] == tr[now].d[2]) //相等时只更新不插入        {            tr[p].w += tr[now].w;            m--;            return;        }        if (tr[now].d[D] >= tr[p].d[D])        {            if (tr[p].r == 0)            {                tr[p].r = now;                return;            }            else            {                p = tr[p].r;            }        }        else        {            if (tr[p].l == 0)            {                tr[p].l = now;                return;            }            else            {                p = tr[p].l;            }        }        D = (D + 1) % 3;    }}int query(int now, int x1, int y1, int z1, int x2, int y2, int z2){    // 当前节点范围与查询矩形不相交    if (!now || tr[now].mx[0] < x1 || tr[now].mn[0] > x2 || tr[now].mx[1] < y1 || tr[now].mn[1] > y2 || tr[now].mx[2] < z1 || tr[now].mn[2] > z2)    {        return 0;    }    // 当前节点范围包含查询矩形    if (tr[now].mx[0] <= x2 && tr[now].mn[0] >= x1 && tr[now].mx[1] <= y2 && tr[now].mn[1] >= y1 && tr[now].mx[2] <= z2 && tr[now].mn[2] >= z1)    {        return tr[now].sum;    }    int ret = 0;    // 查询矩形包含当前节点    if (tr[now].d[0] >= x1 && tr[now].d[0] <= x2 && tr[now].d[1] >= y1 && tr[now].d[1] <= y2 && tr[now].d[2] >= z1 && tr[now].d[2] <= z2)    {        ret += tr[now].w;    }    ret += query(tr[now].l, x1, y1, z1, x2, y2, z2) + query(tr[now].r, x1, y1, z1, x2, y2, z2);    return ret;}int main(){#ifdef LOCAL    freopen("data.in", "r", stdin);    freopen("data.out", "w", stdout);#endif    int t;    scanf("%d", &t);    while (t--)    {        scanf("%d", &n);        int op;        m = 0;        int mod = 10000;        root = 1;        while (n--)        {            scanf("%d", &op);            if (op == 1)            {                int x, y, z;                scanf("%d %d %d", &x, &y, &z);                tr[++m] = data(x, y, z, 1);                kdinsert(m);                if (m % mod == 0) //每mod次重建KD树                {                    root = build(1, m, 0);                }            }            else            {                int x1, y1, z1, x2, y2, z2;                scanf("%d %d %d %d %d %d", &x1, &y1, &z1, &x2, &y2, &z2);                int ans = query(root, x1, y1, z1, x2, y2, z2);                printf("%d\n", ans);            }        }    }    return 0;}