题意：
有N个小松鼠，它们的家用一个点x,y表示，两个点的距离定义为：点(x,y)和它周围的8个点即上下左右四个点和对角的四个点，距离为1。现在N个松鼠要走到一个松鼠家去，求走过的最短距离。
n <= 1e5;
思路：
题意描述的是切比雪夫距离，就是两点之间的距离为max（dx，dy）。要求所有点的话，用曼哈顿距离配上前缀和能比较快得求出来。
所以要把切比雪夫距离转化为曼哈顿距离。
曼哈顿距离通过对x，y坐标分别排序求前缀和，可以O（n）得出所有点的曼哈顿距离前缀和。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; typedef pair<ll,int>pli; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl '\n' //#define R register #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1e9+7; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } /*-----------------------showtime----------------------*/ const int maxn = 1e5+9; struct node { ll x,y; int id; }p[maxn]; bool cmpx(node a, node b){ return a.x < b.x; } bool cmpy(node a,node b){ return a.y < b.y; } ll sum[maxn],sx[maxn],sy[maxn]; int main(){ int n; scanf("%d", &n); for(int i=1; i<=n; i++){ ll x,y; scanf("%lld%lld", &x, &y); p[i].x = x+y; p[i].y = x-y; p[i].id = i; } sort(p+1,p+1+n,cmpx); for(int i=1; i<=n; i++) sx[i] = sx[i-1] + p[i].x; for(int i=1 ;i<=n; i++){ sum[p[i].id] = p[i].x*(i-1) - sx[i-1] + sx[n] - sx[i] - p[i].x*(n-i); } sort(p+1,p+1+n,cmpy); for(int i=1; i<=n; i++) sy[i] = sy[i-1] + p[i].y; ll ans = inff; for(int i=1 ;i<=n; i++){ ll tmp = p[i].y*(i-1) - sy[i-1] + sy[n] - sy[i] - p[i].y*(n-i); ans = min(ans, sum[p[i].id] + tmp); } printf("%lld\n", ans/2); return 0; }