HDU 2966 In case of failure (k-d树)

题意：给出n个点，找出每个点距离他们最近的点的距离的平方。

思路：k-d树的模板题，先将所有点建立好k-d树，然后在依次求一下距离他们最近的点，如果找到的点是他们自己的时候设为距离为无穷大即可。

///k-d树#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>typedef long long ll;const int maxn = 2 * 1e5 + 10;const ll INF = 3 * 1e18;const double eps = 1e-6;using namespace std;int k, n, nodecnt, t; ///点的维数,个数struct point {    ll data[5];    int split; ///分裂维    void input() { for(int i = 0; i < k; i++) scanf("%lld", &data[i]); }    bool operator < (point p) const {        int i = split;        if(data[i] != p.data[i]) return data[i] < p.data[i];        for(int j = 0; j < k; j++) {            if(data[j] == p.data[j]) continue;            return data[j] < p.data[j];        }        return false;    }} p[maxn], ot[maxn];struct node {    int son[2];    point p;    int split; ///分裂维} kd[maxn];///建立k-d树void build(int o, int l, int r) {    if(r < l) return ;    int nn = r - l + 1, sp;    double EX[5], EXX[5];    memset(EX, 0, sizeof EX);     ///E(X)    memset(EXX, 0, sizeof EXX);   ///E(X^2)    for(int i = l; i <= r; i++) {        for(int j = 0; j < k; j++) {            EX[j] += (double)p[i].data[j] / nn;            EXX[j] += (double)p[i].data[j] * p[i].data[j] / nn;        }    }    ///D(X) = E(X^2) - E(X)^2，将方差最大的维作为分裂维    double dx = -1;    for(int i = 0; i < k; i++) {        double d = EXX[i] - EX[i] * EX[i];        if(dx < d) { dx = d; sp = i; }    }    kd[o].split = sp;    for(int i = l; i <= r; i++) {        p[i].split = sp;    }    int mid = (l + r) >> 1;    ///sort(p + l, p + r + 1);    nth_element (p + l, p + mid, p + r + 1);  ///O(n)    for(int i = 0; i < k; i++) {        kd[o].p.data[i] = p[mid].data[i];    }    if(mid - 1 >= l) {        nodecnt++; kd[o].son[0] = nodecnt;        build(kd[o].son[0], l, mid - 1);    }    if(mid + 1 <= r) {        nodecnt++; kd[o].son[1] = nodecnt;        build(kd[o].son[1], mid + 1, r);    }}///两点的距离平方ll dis(point a, point b) {    ll d = 0;    for(int i = 0; i < k; i++) {        d += (a.data[i] - b.data[i]) * (a.data[i] - b.data[i]);    }    return d;}///寻找距离点p最近的点ll mindistance(int o, point po) {    if(!o) return INF;    int s = 0;    ///寻找进入的子树    if(kd[o].p.data[kd[o].split] != po.data[kd[o].split]) {        s = kd[o].p.data[kd[o].split] > po.data[kd[o].split] ? 0 : 1;    } else {        for(int i = 0; i < k; i++) {            if(kd[o].p.data[i] == po.data[i]) continue;            s = kd[o].p.data[i] > po.data[i] ? 0 : 1;            break;        }    }    ///查询点的位置    int nxt = kd[o].son[s];    int other = kd[o].son[s ^ 1];    ll di = dis(kd[o].p, po), dr = 0;    if(!di) di = INF;    ll dl = mindistance(nxt, po);    if(dl) di = min(di, dl);    if(!di) di = INF;    double g = sqrt(di);        ///是否要进入另一棵子树    ll f = s ? -1 : 1;    ll lo = kd[o].p.data[kd[o].split]; ///当前节点分裂维的坐标    double lp = po.data[kd[o].split] + f * sqrt(di);    if(s == 0 && lp + eps > lo) dr = mindistance(other, po);    if(s == 1 && lp - eps < lo) dr = mindistance(other, po);    if(di && dr) di = min(di, dr);    else if(!di) di = dr;    if(!di) di = INF;    return di;}int main() {    k = 2;    scanf("%d", &t);    while(t--) {        scanf("%d", &n);  nodecnt = 1;        for(int i = 0; i < maxn; i++) {            kd[i].son[0] = kd[i].son[1] = 0;        }        for(int i = 0; i < n; i++) {            p[i].input();            ot[i].data[0] = p[i].data[0];            ot[i].data[1] = p[i].data[1];        }        build(1, 0, n - 1);        for(int i = 0; i < n; i++) {            printf("%lld\n", mindistance(1, ot[i]));        }    }    return 0;}