BZOJ 28 (kdtree)

题意：给出n个点，接下来m个操作，每次插入一个点，或者询问离询问点的最近曼哈顿距离。

直接暴力插点询问即可。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <stack>
#define Clear(x,y) memset (x,y,sizeof(x))
#define Close() ios::sync_with_stdio(0)
#define Open() freopen ("more.in", "r", stdin)
#define get_min(a,b) a = min (a, b)
#define get_max(a,b) a = max (a, b);
#define fi first
#define se second
#define pii pair<int, int>
#define pli pair<long long, int>
#define pb push_back
#define mod 1000000007
template <class T>
inline bool scan (T &ret) {
    char c;
    int sgn;
    if (c = getchar(), c == EOF) return 0; //EOF
    while (c != '-' && (c < '0' || c > '9') ) c = getchar();
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return 1;
}
using namespace std;
#define maxn 500005

struct node {
    int d[2], l, r;//节点的点的坐标 左右孩子
    int Max[2], Min[2];//节点中点x的最值 y的最值
}tree[maxn<<1], tmp;
int n, m;
int root, cmp_d;

bool cmp (const node &a, const node &b) {
    return a.d[cmp_d] < b.d[cmp_d] || (a.d[cmp_d] == b.d[cmp_d] &&
            a.d[cmp_d^1] < b.d[cmp_d^1]);
}

void push_up (int p, int pp) {
    get_min (tree[p].Min[0], tree[pp].Min[0]);
    get_min (tree[p].Min[1], tree[pp].Min[1]);
    get_max (tree[p].Max[0], tree[pp].Max[0]);
    get_max (tree[p].Max[1], tree[pp].Max[1]);
}

int build_tree (int l, int r, int D) {
    int mid = (l+r)>>1;
    tree[mid].l = tree[mid].r = 0;
    cmp_d = D;
    nth_element (tree+l+1, tree+mid+1, tree+1+r, cmp);
    //按照cmp把第mid元素放在中间 比他小的放左边 比他大的放右边
    tree[mid].Max[0] = tree[mid].Min[0] = tree[mid].d[0];
    tree[mid].Max[1] = tree[mid].Min[1] = tree[mid].d[1];
    if (l != mid) tree[mid].l = build_tree (l, mid-1, D^1);
    if (r != mid) tree[mid].r = build_tree (mid+1, r, D^1);
    if (tree[mid].l) push_up (mid, tree[mid].l);
    if (tree[mid].r) push_up (mid, tree[mid].r);
    return mid;
}

void insert (int now) {
    int D = 0, p = root;
    while (1) {
        push_up (p, now);//先更新p节点
        if (tree[now].d[D] >= tree[p].d[D]) {
            if (!tree[p].r) {
                tree[p].r = now;
                return ;
            }
            else p = tree[p].r;
        }
        else {
            if (!tree[p].l) {
                tree[p].l = now;
                return ;
            }
            else p = tree[p].l;
        }
        D ^= 1;
    }
    return ;
}

#define INF 1e9
int ans, x, y;
int dis (int p, int x, int y) {//点(x,y)在p的管辖范围内的可能最小值
    int ans = 0;
    if (x < tree[p].Min[0]) ans += tree[p].Min[0]-x;
    if (x > tree[p].Max[0]) ans += x-tree[p].Max[0];
    if (y < tree[p].Min[1]) ans += tree[p].Min[1]-y;
    if (y > tree[p].Max[1]) ans += y-tree[p].Max[1];
    return ans;
}

void query (int p) {
    int dl = INF, dr = INF, d0;
    d0 = abs (tree[p].d[0]-x) + abs (tree[p].d[1]-y);//初始答案
    get_min (ans, d0);
    if (tree[p].l) dl = dis (tree[p].l, x, y);
    if (tree[p].r) dr = dis (tree[p].r, x, y);
    if (dl < dr) {
        if (dl < ans) query (tree[p].l);
        if (dr < ans) query (tree[p].r);
    }
    else {
        if (dr < ans) query (tree[p].r);
        if (dl < ans) query (tree[p].l);
    }
}

int main () {
    //Open ();
    while (scanf ("%d%d", &n, &m) == 2) {
        for (int i = 1; i <= n; i++) {
            scan (tree[i].d[0]);
            scan (tree[i].d[1]);
        }
        root = build_tree (1, n, 0);
        for (int i = 1; i <= m; i++) { 
            int op;
            scan (op); scan (x); scan (y);
            if (op == 1) {
                n++; tree[n].d[0] = x, tree[n].d[1] = y;
                tree[n].Max[0] = tree[n].Min[0] = x;
                tree[n].Max[1] = tree[n].Min[1] = y;
                insert (n);
            }
            else {
                ans = INF;
                query (root);
                printf ("%d\n", ans);
            }
        }
    }
    return 0;
}