hdu 5126 stars cdq分治套cdq分治+树状数组

题目链接

给n个操作， 第一种是在x, y, z这个点+1. 第二种询问(x1, y1, z1). (x2, y2, z2)之间的总值。

用一次cdq分治可以将三维变两维， 两次的话就变成一维了， 然后最后一维用树状数组维护。 对于每个询问， 相当于将它拆成8个点。

注意第二次cdq分治的时候l可能小于r。 所以这里的return条件是l <= r而不是l == r。 找了好久...

#include <bits/stdc++.h>
using namespace std;
#define pb(x) push_back(x)
#define mem(a) memset(a, 0, sizeof(a))
vector <int> b;
const int maxn = 5e4+5;
struct node
{
    int x1, y1, z1, x2, y2, z2;
    int sign, id;
    node(){}
    node(int x, int y, int z, int sign, int id):x1(x), y1(y), z1(z), sign(sign), id(id){}
}a[maxn], s[maxn*8], ss[maxn*8], c[maxn*8];
bool cmpx(node a, node b)
{
    if(a.x1 == b.x1)
        return a.id < b.id;
    return a.x1 < b.x1;
}
bool cmpy(node a, node b)
{
    if(a.y1 == b.y1)
        return a.id < b.id;
    return a.y1 < b.y1;
}
int len, sum[maxn*8], ans[maxn];
int lowbit(int x) {return x&(-x);}
void update(int x, int val) {
    while(x <= len) {
        sum[x] += val;
        x += lowbit(x);
    }
}
int query(int x, int ret = 0)
{
    while(x) {
        ret += sum[x];
        x -= lowbit(x);
    }
    return ret;
}
void cdqy(int l, int r)
{
    if(l >= r)
        return ;
    int m = l+r>>1, cnt = 0;
    cdqy(l, m);
    cdqy(m+1, r);
    for(int i = l; i <= m; i++) {
        if(s[i].sign == 1) {
            ss[cnt++] = s[i];
        }
    }
    for(int i = m+1; i <= r; i++) {
        if(s[i].sign != 1) {
            ss[cnt++] = s[i];
        }
    }
    sort(ss, ss+cnt, cmpy);
    for(int i = 0; i < cnt; i++) {
        if(ss[i].sign == 1) {
            update(ss[i].z1, 1);
        } else if(ss[i].sign == 2) {
            ans[ss[i].id] += query(ss[i].z1);
        } else {
            ans[ss[i].id] -= query(ss[i].z1);
        }
    }
    for(int i = 0; i < cnt; i++) {
        if(ss[i].sign == 1) {
            update(ss[i].z1, -1);
        }
    }
}
void cdqx(int l, int r)
{
    if(l == r)
        return ;
    int m = l+r>>1, top = 0;
    cdqx(l, m);
    cdqx(m+1, r);
    for(int i = l; i <= m; i++) {
        if(c[i].sign == 1) {
            s[++top] = c[i];
        }
    }
    for(int i = m+1; i <= r; i++) {
        if(c[i].sign != 1) {
            s[++top] = c[i];
        }
    }
    sort(s+1, s+1+top, cmpx);
    cdqy(1, top);
}
void solve(int n, int num = 0)
{
    sort(b.begin(), b.end());
    b.erase(unique(b.begin(), b.end()), b.end());
    len = b.size();
    for(int i = 1; i <= n; i++) {
        if(a[i].sign == 1) {
            c[++num] = a[i];
        } else {
            c[++num] = node(a[i].x2, a[i].y2, a[i].z2, 2, a[i].id);
            c[++num] = node(a[i].x1-1, a[i].y1-1, a[i].z1-1, 3, a[i].id);
            c[++num] = node(a[i].x2, a[i].y2, a[i].z1-1, 3, a[i].id);
            c[++num] = node(a[i].x2, a[i].y1-1, a[i].z2, 3, a[i].id);
            c[++num] = node(a[i].x1-1, a[i].y2, a[i].z2, 3, a[i].id);
            c[++num] = node(a[i].x1-1, a[i].y1-1, a[i].z2, 2, a[i].id);
            c[++num] = node(a[i].x1-1, a[i].y2, a[i].z1-1, 2, a[i].id);
            c[++num] = node(a[i].x2, a[i].y1-1, a[i].z1-1, 2, a[i].id);
        }
    }
    for(int i = 1; i <= num; i++) {
        c[i].z1 = lower_bound(b.begin(), b.end(), c[i].z1)-b.begin()+1;
    }
    cdqx(1, num);
}
int main()
{
    int t, n;
    cin>>t;
    while(t--) {
        cin>>n;
        mem(sum);
        mem(ans);
        for(int i = 1; i <= n; i++) {
            scanf("%d%d%d%d", &a[i].sign, &a[i].x1, &a[i].y1, &a[i].z1);
            if(a[i].sign == 2) {
                scanf("%d%d%d", &a[i].x2, &a[i].y2, &a[i].z2);
                b.pb(a[i].z2);
                b.pb(a[i].z1-1);
            } else {
                b.pb(a[i].z1);
            }
            a[i].id = i;
        }
        solve(n);
        for(int i = 1; i <= n; i++) {
            if(a[i].sign == 2) {
                printf("%d
", ans[i]);
            }
        }
    }
    return 0;
}