BZOJ3170: [Tjoi 2013]松鼠聚会

3170: [Tjoi 2013]松鼠聚会

Time Limit: 10 Sec  Memory Limit: 128 MB

Description

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。

Input

第一行给出数字N，表示有多少只小松鼠。0<=N<=10^5
下面N行，每行给出x,y表示其家的坐标。
-10^9<=x,y<=10^9

Output

表示为了聚会走的路程和最小为多少。

Sample Input

6
-4 -1
-1 -2
2 -4
0 2
0 3
5 -2

Sample Output

20

HINT

 

Source

转换一下就A了，，，

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<vector>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pr;
const double pi=acos(-1);
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,n,a) for(int i=n;i>=a;i--)
#define Rep(i,u) for(int i=head[u];i;i=Next[i])
#define clr(a) memset(a,0,sizeof(a))
#define pb push_back
#define mp make_pair
#define fi first
#define sc second
#define pq priority_queue
#define pqb priority_queue <int, vector<int>, less<int> >
#define pqs priority_queue <int, vector<int>, greater<int> >
#define vec vector
ld eps=1e-9;
ll pp=1000000007;
ll mo(ll a,ll pp){if(a>=0 && a<pp)return a;a%=pp;if(a<0)a+=pp;return a;}
ll powmod(ll a,ll b,ll pp){ll ans=1;for(;b;b>>=1,a=mo(a*a,pp))if(b&1)ans=mo(ans*a,pp);return ans;}
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
//void add(int x,int y,int z){ v[++e]=y; next[e]=head[x]; head[x]=e; cost[e]=z; }
int dx[5]={0,-1,1,0,0},dy[5]={0,0,0,-1,1};
ll read(){ ll ans=0; char last=' ',ch=getchar();
while(ch<'0' || ch>'9')last=ch,ch=getchar();
while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
if(last=='-')ans=-ans; return ans;
}
const int N=100005;
ll X[N],Y[N],A[N],B[N],sA[N],sB[N],ans=1e15;
int main(){
    int n=read(),a,b;
    for (int i=1;i<=n;i++){
        a=read(),b=read();
        X[i]=a+b; Y[i]=a-b;
        A[i]=a+b; B[i]=a-b;
    }
    sort(A+1,A+n+1);
    sort(B+1,B+n+1);
    for (int i=1;i<=n;i++){
        sA[i]=sA[i-1]+A[i]; sB[i]=sB[i-1]+B[i];
    }
    int p,q;
    for (int i=1;i<=n;i++){
        p=lower_bound(A+1,A+n+1,X[i])-A; q=lower_bound(B+1,B+n+1,Y[i])-B;
        ans=min(ans,(X[i]*p-sA[p])+((sA[n]-sA[p])-X[i]*(n-p))+(Y[i]*q-sB[q])+((sB[n]-sB[q])-Y[i]*(n-q)));
    }
    cout<<ans/2;
    return 0;
}