BZOJ-2716-天使玩偶angel-CDQ分治

描述

先给出n个点, 然后有m个操作, (1, x, y) 表示查询离(x, y)最近点的曼哈顿距离, (2, x, y) 表示插入点 (x, y).

分析

• 不会做... 又照着别人的代码打了一遍... CDQ分治总想不到思路
• 比较关键的几个地方是 : 1. 坐标的范围是小于1000000的所以可以用树状数组维护. 2. 距离点(x, y)最近的点和x的方位有四种, 左下左上右下右上, 然后只考虑一个方位, 另外的改变坐标即可. 3. 曼哈顿距离不是欧几里得距离, 是横纵坐标之差绝对值的和. dis({x, y}, {x', y'}) = |x-x'| + |y-y'|. 在只考虑(x', y')在(x, y)的左下方时, 可以去掉绝对值 : dis = x+y - (x'+y'), 使x'+y'最大即可.
• 分治时始终保持横坐标递增, 分治时考虑左边对右边的影响, 上面说了要使dis = x+y - (x'+y')的x'+y'最大, 就按 x' 排序按 y' 用树状数组维护x'+y'的最大值. 在树状数组中查询小于y的最大x'+y'值.

代码

#include 
#include 
using namespace std;

const int maxn = 1000000 + 10;
const int INF = 1000000000;

int max_x;
int ans[maxn];

struct BIT {
	int c[maxn];
	int lowbit(int x) {
		return x & (-x);
	}
	void modify(int x, int d) {
		while(x <= max_x) {
			c[x] = max(c[x], d);
			x += lowbit(x);
		}
	}
	int query(int x) {
		int ret = 0;
		while(x > 0) {
			ret = max(ret, c[x]);
			x -= lowbit(x);
		}
		return ret;
	}
	void clear(int x) {
		while(x <= max_x) {
			c[x] = 0;
			x += lowbit(x);
		}
	}
} bit;

struct Node {
	int x, y, k, id;
	bool operator < (const Node& rhs) const {
		if(x != rhs.x) return x < rhs.x;
		return id < rhs.id;
	}
} A[maxn];

#define a A[i]
#define b A[j]

struct CDQ {
	int n;
	Node T[maxn];
	
	void init(int n) {
		this->n = n;
		sort(A+1, A+n+1);
	}
	
	void solve(int L, int R) {
		if(L >= R) return;
		int M = (L+R) >> 1;
		int i, j, p = L, q = M+1;
		for(i = L; i <= R; i++) if(a.id <= M) T[p++] = a; else T[q++] = a;
		for(i = L; i <= R; i++) A[i] = T[i];
		solve(L, M);
		i = M+1; j = L;
		for(; i <= R; i++) if(a.k == 2) {
			for(; j <= M && b.x <= a.x; j++) if(b.k == 1)
				bit.modify(b.y, b.x+b.y);
			int t = bit.query(a.y);
			if(t) ans[a.id] = min(ans[a.id], a.x+a.y-t);
		}
		for(i = L; i < j; i++) if(a.k == 1)
			bit.clear(a.y);
		solve(M+1, R);
		merge(A+L, A+M+1, A+M+1, A+R+1, T+L);
		for(i = L; i <= R; i++) A[i] = T[i];
	}
	
} cdq;

int main()
{
	int n, m;
	scanf("%d %d", &n, &m);
	for(int i = 1; i <= n; i++) {
		scanf("%d %d", &a.x, &a.y);
		a.x++; a.y++; a.id = i; a.k = 1;
		max_x = max(max_x, max(a.x, a.y));
	}
	for(int i = n+1; i <= n+m; i++) {
		scanf("%d %d %d", &a.k, &a.x, &a.y);
		a.x++; a.y++; a.id = i;
		max_x = max(max_x, max(a.x, a.y));
	}
	max_x++; n += m;
	for(int i = 1; i <= n; i++) ans[i] = INF;
	
	cdq.init(n); cdq.solve(1, n);
	
	for(int i = 1; i <= n; i++) a.x = max_x - a.x;
	cdq.init(n); cdq.solve(1, n);
	
	for(int i = 1; i <= n; i++) a.y = max_x - a.y;
	cdq.init(n); cdq.solve(1, n);
	
	for(int i = 1; i <= n; i++) a.x = max_x - a.x;
	cdq.init(n); cdq.solve(1, n);
	
	for(int i = 1; i <= n; i++)
		if(ans[i] != INF) printf("%d\n", ans[i]);
	return 0;
}