洛谷P4169 天使玩偶 (算竞进阶习题)

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 <bits/stdc++.h>
#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<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
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;
}