D-Distance_2019牛客暑期多校训练营（第八场）
题目链接
题意
1<=nmh,q<=1e5
q个操作
- 1 x y z往坐标里加入一个点
- 2 x y z查询距离该点最近的点的距离(曼哈顿距离)
题解
做法一
将要插入的点保存在一个队列中，当队列中的点数达到一个阈值就把队列中所有点取出，暴力的bfs一次把答案记录在\(dis[getid(x,y,z)]\)中表示距离点\((x,y,z)\)最近的点的距离，查询的时候就取暴力的计算查询点和队列中每个点的距离，再和已经插入的点也就是dis数组取最小值，当阈值取\(\sqrt{nmh}\)时复杂度为\(O(\frac{qnmh}E + qE) = O(nmh + q\sqrt{nmh})\)
做法二
将距离公式\(|x_0-x_i| + |y_0-y_i| + |z_0-z_i|\)的绝对值拆开有八种情况\(\pm(x_0-x_i) \pm (y_0-y_i) \pm (z_0-z_i)\)
这八种情况的最大值就是真正的距离，我们将插入的点分成\((\pm x, \pm y, \pm z)\)八种情况分别插入八个树状数组，树状数组维护\(x_i <= x， y_i <= y， z_i <= z的x+y+z\)的最大值，目的是为了把求最近点的距离转换成求\((x+y+z-x_i-y_i-z_i)\)的最小值，查询的时候对八个树状数组答案取min就行了
代码
做法一
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mx = 1e5+5;
const int mod = 998244353;
const int INF = 0x3f3f3f3f;
const int dir[6][3] = {{0,0,1}, {0,0,-1}, {0,1,0}, {0,-1,0}, {1,0,0}, {-1,0,0}};
int dis[mx];
int n, m, h, q;
struct node {
int x, y, z, step;
};
vector <node> v;
queue <node> Q;
int getid(int x, int y, int z) {
return (x-1)*m*h + (y-1)*h + z;
}
int getdis(int x, int y, int z, int a, int b, int c) {
return abs(x-a) + abs(y-b) + abs(z-c);
}
int main() {
for (int i = 0; i < mx; i++) dis[i] = INF;
scanf("%d%d%d%d", &n, &m, &h, &q);
int sq = sqrt(n*m*h) ;
while (q--) {
int op, x, y, z;
scanf("%d%d%d%d", &op, &x, &y, &z);
if (op == 1) {
v.push_back({x, y, z, 0});
if (v.size() == sq) {
for (int i = 0; i < v.size(); i++) {
Q.push(v[i]);
dis[getid(v[i].x, v[i].y, v[i].z)] = 0;
}
v.clear();
while (!Q.empty()) {
node now = Q.front();
node next;
Q.pop();
for (int i = 0; i < 6; i++) {
next.x = now.x + dir[i][0];
next.y = now.y + dir[i][1];
next.z = now.z + dir[i][2];
if (next.x < 1 || next.x > n || next.y < 1 || next.y > m || next.z < 1 || next.z > h) continue;
next.step = now.step + 1;
if (next.step < dis[getid(next.x, next.y, next.z)]) {
dis[getid(next.x, next.y, next.z)] = next.step;
Q.push(next);
}
}
}
}
} else {
int ans = dis[getid(x, y, z)];
for (int i = 0; i < v.size(); i++) {
ans = min(ans, getdis(x, y, z, v[i].x, v[i].y, v[i].z));
}
printf("%d\n", ans);
}
}
return 0;
}
做法二
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mx = 8e5+5;
const int mod = 998244353;
const int INF = 0x3f3f3f3f;
const int dir[8][3] = {{1,1,1}, {1,1,-1}, {1,-1,1}, {1,-1,-1}, {-1,1,1}, {-1,1,-1}, {-1,-1,1}, {-1,-1,-1}};
int dis[mx];
int n, m, h, q;
int getid(int x, int y, int z) {
return (x-1)*m*h + (y-1)*h + z;
}
struct Bit {
int a[mx];
int lowbit(int x) {
return x & -x;
}
void update(int x, int y, int z) {
if (x < 0) x += n;
if (y < 0) y += m;
if (z < 0) z += h;
for (int i = x; i <= n; i+=lowbit(i))
for (int j = y; j <= m; j+=lowbit(j))
for (int k = z; k <= h; k+=lowbit(k))
a[getid(i,j,k)] = max(a[getid(i,j,k)], x+y+z);
}
int query(int x, int y, int z) {
if (x < 0) x += n;
if (y < 0) y += m;
if (z < 0) z += h;
int ans = 0;
for (int i = x; i > 0; i-=lowbit(i))
for (int j = y; j > 0; j-=lowbit(j))
for (int k = z; k > 0; k-=lowbit(k))
ans = max(a[getid(i,j,k)], ans);
if (ans == 0) return INF;
else return x+y+z-ans;
}
}bit[8];
int main() {
scanf("%d%d%d%d", &n, &m, &h, &q);
n++; m++; h++;
while (q--) {
int op, x, y, z;
scanf("%d%d%d%d", &op, &x, &y, &z);
if (op == 1) {
for (int i = 0; i < 8; i++) {
bit[i].update(x*dir[i][0], y*dir[i][1], z*dir[i][2]);
}
} else {
int ans = INF;
for (int i = 0; i < 8; i++) ans = min(ans, bit[i].query(x*dir[i][0], y*dir[i][1], z*dir[i][2]));
printf("%d\n", ans);
}
}
return 0;
}
