【K-D树 求最近最远距离】BZOJ 1941 [Sdoi2010]Hide and Seek

Problem Description

给你n个点（二维），让你求所有点 到最远点 和最近点 的距离差最小

思路：

自身点不算。
求个距离该点，最近距离。和 距离该点，最远距离即可。
这里的距离指的是 曼哈顿距离

#include<bits/stdc++.h>using namespace std;const int MAX = 500010;const int inf = 0x3f3f3f3f;const int DIM = 2;struct node{    int l, r;    int d[DIM], maxn[DIM], minn[DIM];    inline void maintain()//初始化    {        l = r = 0;        for(int i = 0; i < DIM; i++)            maxn[i] = minn[i] = d[i];    }}tree[MAX];int D;bool operator < (const node &a, const node &b)//重载，从小到大{    return a.d[D] < b.d[D];}inline void Merge(int mid)//归并，向上更新{    int son[2] = {tree[mid].l, tree[mid].r};    for(int i = 0; i < 2; i++)    {        if(!son[i]) continue;        for(int j = 0; j < DIM; j++)        {            tree[mid].maxn[j] = max(tree[mid].maxn[j], tree[son[i]].maxn[j]);            tree[mid].minn[j] = min(tree[mid].minn[j], tree[son[i]].minn[j]);        }    }}int build(int l, int r, int now)//建树{    int mid = (l+r) >> 1;    D = now;    nth_element(tree+l, tree+mid, tree+r+1);//对于now维，以mid为中间，左边都是比它小，右边都是比它大。    tree[mid].maintain();//初始化    if(l < mid) tree[mid].l = build(l, mid-1, (now+1)%DIM);    if(r > mid) tree[mid].r = build(mid+1, r, (now+1)%DIM);    Merge(mid);    return mid;}int ans1, ans2;int partionMin(int o, int k)//求最小可能值{    int red = 0;    for(int i = 0; i < DIM; i++)    {        if(tree[k].d[i] > tree[o].maxn[i]) red += tree[k].d[i] - tree[o].maxn[i];        if(tree[k].d[i] < tree[o].minn[i]) red += tree[o].minn[i] - tree[k].d[i];    }    return red;}int partionMax(int o, int k)//求最大可能值{    int red = 0;    for(int i = 0; i < DIM; i++)    {        red += max(abs(tree[k].d[i] - tree[o].maxn[i]), abs(tree[k].d[i] - tree[o].minn[i]));    }    return red;}void queryMin(int o, int k)//求最小距离{    int dm = abs(tree[k].d[0] - tree[o].d[0]) + abs(tree[k].d[1] - tree[o].d[1]);    if(o == k) dm = inf;//自身点不算。    if(ans1 > dm) ans1 = dm;    int dl = tree[o].l ? partionMin(tree[o].l, k) : inf;    int dr = tree[o].r ? partionMin(tree[o].r, k) : inf;    if(dl < dr)    {        if(dl < ans1) queryMin(tree[o].l, k);        if(dr < ans1) queryMin(tree[o].r, k);    }    else    {        if(dr < ans1) queryMin(tree[o].r, k);        if(dl < ans1) queryMin(tree[o].l, k);    }}void queryMax(int o, int k)//求最大距离{    int dm = abs(tree[k].d[0] - tree[o].d[0]) + abs(tree[k].d[1] - tree[o].d[1]);    if(ans2 < dm) ans2 = dm;    int dl = tree[o].l ? partionMax(tree[o].l, k) : 0;    int dr = tree[o].r ? partionMax(tree[o].r, k) : 0;    if(dl > dr)    {        if(dl > ans2) queryMax(tree[o].l, k);        if(dr > ans2) queryMax(tree[o].r, k);    }    else    {        if(dr > ans2) queryMax(tree[o].r, k);        if(dl > ans2) queryMax(tree[o].l, k);    }}int main(){    int n, i, j;    while(~scanf("%d", &n))    {        for(i = 1; i <= n; i++)            for(j = 0; j < DIM; j++)            scanf("%d", &tree[i].d[j]);        int root = build(1, n, 0);        int Ans;        for(i = 1; i <= n; i++)        {            ans1 = inf;            ans2 = 0;            queryMin(root, i);            queryMax(root, i);            Ans = min(Ans, ans2 - ans1);        }        printf("%d\n", Ans);    }}