【题目链接】
https://www.lydsy.com/JudgeOnline/problem.php?id=3170
【题解】
首先将切比雪夫距离转换为曼哈顿距离。
把每个点的坐标变为
后曼哈顿距离等于之前的切比雪夫距离。
证明:
设切比雪夫距离为
,曼哈顿距离为
,两个点的坐标为
。
那么有
设
那么有
展开得
所以
转换为曼哈顿距离之后,就可以横纵坐标分开算+扫描线了。
时间复杂度
【代码】
/* - - - - - - - - - - - - - - -
User : VanishD
problem : [bzoj3170]
Points : Chebyshev Distance
- - - - - - - - - - - - - - - */
# include <bits/stdc++.h>
# define ll long long
# define N 100100
using namespace std;
const int inf = 0x3f3f3f3f, INF = 0x7fffffff;
const ll infll = 0x3f3f3f3f3f3f3f3fll, INFll = 0x7fffffffffffffffll;
int read(){
int tmp = 0, fh = 1; char ch = getchar();
while (ch < '0' || ch > '9'){ if (ch == '-') fh = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9'){ tmp = tmp * 10 + ch - '0'; ch = getchar(); }
return tmp * fh;
}
struct Node{
ll x, y;
int id;
}p[N];
ll ans[N];
int n;
bool cmpx(Node x, Node y){ return x.x < y.x; }
bool cmpy(Node x, Node y){ return x.y < y.y; }
int main(){
// freopen(".in", "r", stdin);
// freopen(".out", "w", stdout);
n = read();
for (int i = 1; i <= n; i++){
ll x = read(), y = read();
p[i].x = x + y, p[i].y = x - y;
p[i].id = i;
}
sort(p + 1, p + n + 1, cmpx);
ll dis0 = 0, dis1 = 0, cnt0 = n - 1, cnt1 = 0;
for (int i = 2; i <= n; i++) dis0 += p[i].x - p[1].x;
for (int i = 1; i <= n; i++){
ans[p[i].id] = dis0 + dis1;
if (i != n){
dis0 = dis0 - 1ll * cnt0 * (p[i + 1].x - p[i].x);
cnt0--; cnt1++;
dis1 = dis1 + 1ll * cnt1 * (p[i + 1].x - p[i].x);
}
}
sort(p + 1, p + n + 1, cmpy);
dis0 = 0, dis1 = 0, cnt0 = n - 1, cnt1 = 0;
for (int i = 2; i <= n; i++) dis0 += p[i].y - p[1].y;
for (int i = 1; i <= n; i++){
ans[p[i].id] += dis0 + dis1;
if (i != n){
dis0 = dis0 - 1ll * cnt0 * (p[i + 1].y - p[i].y);
cnt0--; cnt1++;
dis1 = dis1 + 1ll * cnt1 * (p[i + 1].y - p[i].y);
}
}
ll final = INFll;
for (int i = 1; i <= n; i++)
final = min(final, ans[i] / 2);
printf("%lld\n", final);
return 0;
}