P3964 [TJOI2013]松鼠聚会

题意：给出 $n$个点 $(x_i,y_i)$，找到某个点，使得所有点到该点的切比雪夫距离和最小。

• 一、
  $A(x_1, y_1), B(x_2, y_2)$
  三种距离：
  ①欧几里得距离： $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
  ②曼哈顿距离： $|x_2-x_1|+|y_2-y_1|$
  ③切比雪夫距离： $max(|x_2-x_1|,|y_2-y_1|)$
• 二、
  相同距离
  曼哈顿距离 $(x,y)$转化为切比雪夫距离： $(x+y,x-y)$
  切比雪夫距离 $(x,y)$转化为曼哈顿距离： $(\displaystyle \frac{x + y}{2}, \frac{x-y}{2})$
• 三、
  对于曼哈顿距离：
  当前枚举到点 $\ i:(x_i,y_i)$，那么所有点到该点的距离： $\displaystyle \sum^{n}_{j=1}{|x_i-x_j|}+\displaystyle \sum^{n}_{j=1}{|y_i-y_j|}$

1. 如果我们将所有横坐标按照升序排序
   $dis_x=\displaystyle \sum^{i}_{j=1}{(x_i-x_j)}+\displaystyle \sum^{n}_{j=i+1}{(x_j-x_i)}$
2. 同理，如果我们将所有纵坐标按照升序排序
   $dis_y=\displaystyle \sum^{i}_{j=1}{(y_i-y_j)}+\displaystyle \sum^{n}_{j=i+1}{(y_j-y_i)}$
   发现了啥？这不是前缀和吗？
   $dis_x=(i*x_i+prefix[x_i])+(prefix[x_n]-prefix[x_i]+(n-i)*x_i)$
   $dis_y=(i*y_i+prefix[y_i])+(prefix[y_n]-prefix[y_i]+(n-i)*y_i)$

由此可得AC代码~

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f3f3f
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

ll read()
{
    ll x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') f = -f; c = getchar(); }
    while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}

const int maxN = 100005;

struct node{
    ll x, y;
    int id;
    node() {}
    node(ll a, ll b, int c) : x(a), y(b), id(c) {}
}mp[maxN];

bool cmp_x(node n1, node n2) { return n1.x < n2.x; }
bool cmp_y(node n1, node n2) { return n1.y < n2.y; }

ll n;

ll ans[maxN];
ll sum[maxN];

int main()
{
    n = read();
    for(int i = 1; i <= n; ++ i )
    {
        mp[i].x = read(); mp[i].y = read();
        ll xx = mp[i].x + mp[i].y, yy = mp[i].x - mp[i].y;
        mp[i] = node(xx, yy, i);
    }
    sort(mp + 1, mp + n + 1, cmp_x);
    for(int i = 1; i <= n; ++ i )
        sum[i] = sum[i - 1] + mp[i].x;
    for(int i = 1; i <= n; ++ i )
        ans[mp[i].id] = (i * mp[i].x - sum[i] + sum[n] - sum[i] - (n - i) * mp[i].x);
    sort(mp + 1, mp + n + 1, cmp_y);
    for(int i = 1; i <= n; ++ i )
        sum[i] = sum[i - 1] + mp[i].y;
    for(int i = 1; i <= n; ++ i )
        ans[mp[i].id] += (i * mp[i].y - sum[i] + sum[n] - sum[i] - (n - i) * mp[i].y);
    ll res = ans[1];
    for(int i = 2; i <= n; ++ i )
        res = min(res, ans[i]);
    cout << (res >> 1) << endl;
    return 0;
}