CDQ分治

分成四个方向讨论最小值，把所有坐标全部离线处理。

把左边按x轴排序，保证x的顺序，然后树状数组维护每个方向需要的最值。。

然后CDQ分治。。必须手动撤销树状数组的修改，保证分治的时间复杂度。

#include 
    
     #define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)using namespace std;typedef long long ll;inline int lowbit(int x){ return x & (-x); }inline int read(){    int X = 0, w = 0; char ch = 0;    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();    return w ? -X : X;}inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }inline int lcm(int a, int b){ return a / gcd(a, b) * b; }template
     
      inline T max(T x, T y, T z){ return max(max(x, y), z); }template
      
       inline T min(T x, T y, T z){ return min(min(x, y), z); }template
       
        inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}const int N = 1000005;int n, m, tot, t[N], k, ans[N];struct rec {    int x, y, z;    bool operator < (const rec &rhs) const {        return x < rhs.x;    }}a[N], b[N];inline void add(int index, int val){    for(; index < tot; index += lowbit(index))        t[index] = max(t[index], val);}inline int query(int index){    int ret = -(1 << 30);    for(; index; index -= lowbit(index))        ret = max(ret, t[index]);    return ret;}inline void calc1(){    for(int i = 1; i <= k; i ++){        int y = b[i].y;        int temp = b[i].x + b[i].y;        if(a[b[i].z].z == 1) add(y, temp);        else ans[b[i].z] = min(ans[b[i].z], abs(temp - query(y)));    }    for(int i = 1; i <= k; i ++){        int y = b[i].y;        if(a[b[i].z].z == 1){            for(int j = y; j < tot; j += lowbit(j)) t[j] = -(1 << 30);        }    }}inline void calc2(){    for(int i = 1; i <= k; i ++){        int y = b[i].y;        int temp = b[i].x - b[i].y;        if(a[b[i].z].z == 1) add(tot - y, temp);        else ans[b[i].z] = min(ans[b[i].z], abs(temp - query(tot - y)));    }    for(int i = 1; i <= k; i ++){        int y = b[i].y;        if(a[b[i].z].z == 1){            for(int j = tot - y; j < tot; j += lowbit(j)) t[j] = -(1 << 30);        }    }}inline void calc3(){    for(int i = k; i >= 1; i --){        int y = b[i].y;        int temp = -b[i].x - b[i].y;        if(a[b[i].z].z == 1) add(tot - y, temp);        else ans[b[i].z] = min(ans[b[i].z], abs(temp - query(tot - y)));    }    for(int i = k; i >= 1; i --){        int y = b[i].y;        if(a[b[i].z].z == 1){            for(int j = tot - y; j < tot; j += lowbit(j)) t[j] = -(1 << 30);        }    }}inline void calc4(){    for(int i = k; i >= 1; i --){        int y = b[i].y;        int temp = -b[i].x + b[i].y;        if(a[b[i].z].z == 1) add(y, temp);        else ans[b[i].z] = min(ans[b[i].z], abs(temp - query(y)));    }    for(int i = k; i >= 1; i --){        int y = b[i].y;        if(a[b[i].z].z == 1){            for(int j = y; j < tot; j += lowbit(j)) t[j] = -(1 << 30);        }    }}void cdqDiv(int l, int r){    if(l == r) return;    int mid = (l + r) >> 1;    cdqDiv(l, mid), cdqDiv(mid + 1, r);    k = 0;    for(int i = l; i <= r; i ++){        if((i <= mid && a[i].z == 1) || (i > mid && a[i].z == 2))            b[++k] = a[i], b[k].z = i;    }    sort(b + 1, b + k + 1);    calc1(), calc2();    calc3(), calc4();}int main(){    full(t, 0xcf), full(ans, INF);    n = read(), m = read();    m += n;    for(int i = 1; i <= n; i ++){        a[i].x = read(), a[i].y = read() + 1, a[i].z = 1;    }    for(int i = n + 1; i <= m; i ++){        a[i].z = read(), a[i].x = read(), a[i].y = read() + 1;    }    for(int i = 1; i <= m; i ++) tot = max(tot, a[i].y);    tot ++;    cdqDiv(1, m);    for(int i = 1; i <= m; i ++){        if(a[i].z == 2) printf("%d\n", ans[i]);    }    return 0;}