这天，SJY显得无聊。在家自己玩。在一个棋盘上，有N个黑色棋子。他每次要么放到棋盘上一个黑色棋子，要么放上一个白色棋子，如果是白色棋子，他会找出距离这个白色棋子最近的黑色棋子。此处的距离是 曼哈顿距离 即(|x1-x2|+|y1-y2|) 。现在给出N<=500000个初始棋子。和M<=500000个操作。对于每个白色棋子，输出距离这个白色棋子最近的黑色棋子的距离。同一个格子可能有多个棋子。

对于每个T=2 输出一个最小距离

2 3
1 1
2 3
2 1 2
1 3 3
2 4 2

1
2

kdtree可以过

#include<cstdio>
#include<algorithm>
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 17, stdin), p1 == p2) ? EOF : *p1++)
using namespace std;
const int MAXN = 6 * 1e5 + 10, INF = 1e9 + 10;
const double delat = 0.60;
char buf[1 << 17], *p1 = buf, *p2 = buf;
inline int read() {char c = getchar(); int x = 0, f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();return x * f;
}
int N, M, WD, root, Ans;
inline int abs(int x) {return x < 0 ? -x : x;
}
#define ls(x) T[x].ls
#define rs(x) T[x].rs
struct Point {int x[2];bool operator < (const Point rhs) const {return x[WD] < rhs.x[WD];}
}p[MAXN];
struct Node {int ls, rs, siz, mi[2], mx[2];Point tp;
}T[MAXN];
int rub[MAXN], top, cur;
int NewNode() {return top ? rub[top--] : ++cur;//tag
}
void update(int k) {T[k].siz = T[ls(k)].siz + T[rs(k)].siz + 1;for(int i = 0; i <= 1; i++) {T[k].mi[i] = T[k].mx[i] = T[k].tp.x[i];if(ls(k)) T[k].mi[i] = min(T[k].mi[i], T[ls(k)].mi[i]), T[k].mx[i] = max(T[k].mx[i], T[ls(k)].mx[i]);if(rs(k)) T[k].mi[i] = min(T[k].mi[i], T[rs(k)].mi[i]), T[k].mx[i] = max(T[k].mx[i], T[rs(k)].mx[i]);}
}
int Build(int l, int r, int wd) {if(l > r) return 0;int k = NewNode(), mid = l + r >> 1;WD = wd, nth_element(p + l, p + mid, p + r + 1);T[k].tp = p[mid];T[k].ls = Build(l, mid - 1, wd ^ 1);T[k].rs = Build(mid + 1, r, wd ^ 1);update(k);return k;
}
inline void Apart(int k, int num) {if(T[k].ls) Apart(ls(k), num);p[num + T[ls(k)].siz + 1] = T[k].tp, rub[++top] = k;if(T[k].rs) Apart(rs(k), num + T[ls(k)].siz + 1);
}
inline int check(int &k, int wd) {if(T[k].siz * delat < T[ls(k)].siz || T[k].siz * delat < T[rs(k)].siz) Apart(k, 0), k = Build(1, T[k].siz, wd);
}
void Insert(Point a, int &k, int wd) {if(k == 0) {k = NewNode(); T[k].tp = a; update(k); return ;}if(a.x[wd] < T[k].tp.x[wd]) Insert(a, ls(k), wd ^ 1);else Insert(a, rs(k), wd ^ 1);update(k); check(k, wd);
}
inline int dis(Point a, Point b) {return abs(a.x[0] - b.x[0]) + abs(a.x[1] - b.x[1]);
}
inline int Manha(Point a, int b) {int rt = 0;for(int i = 0; i <= 1; i++) rt += max(0, a.x[i] - T[b].mx[i]) + max(0, T[b].mi[i] - a.x[i]);return rt;
}
int Query(Point a, int k) {Ans = min(Ans, dis(T[k].tp, a));int disl = INF, disr = INF;if(ls(k)) disl = Manha(a, T[k].ls);if(rs(k)) disr = Manha(a, T[k].rs);if(disl < disr) {if(disl < Ans) Query(a, ls(k));if(disr < Ans) Query(a, rs(k));}else {if(disr < Ans) Query(a, rs(k));if(disl < Ans) Query(a, ls(k));}
}
int main() {#ifdef WIN32freopen("a.in", "r", stdin);#endifN = read(); M = read();for(int i = 1; i <= N; i++) p[i].x[0] = read(), p[i].x[1] = read();root = Build(1, N, 0);while(M--) {int opt = read(), x = read(), y = read();if(opt == 1) Insert((Point){x, y}, root, 0);else Ans = INF + 1, Query((Point){x, y}, root), printf("%d\n", Ans);}return 0;
}