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,,,

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<cstdlib>
 7 #include<vector>
 8 using namespace std;
 9 typedef long long ll;
10 typedef long double ld;
11 typedef pair<int,int> pr;
12 const double pi=acos(-1);
13 #define rep(i,a,n) for(int i=a;i<=n;i++)
14 #define per(i,n,a) for(int i=n;i>=a;i--)
15 #define Rep(i,u) for(int i=head[u];i;i=Next[i])
16 #define clr(a) memset(a,0,sizeof(a))
17 #define pb push_back
18 #define mp make_pair
19 #define fi first
20 #define sc second
21 #define pq priority_queue
22 #define pqb priority_queue <int, vector<int>, less<int> >
23 #define pqs priority_queue <int, vector<int>, greater<int> >
24 #define vec vector
25 ld eps=1e-9;
26 ll pp=1000000007;
27 ll mo(ll a,ll pp){if(a>=0 && a<pp)return a;a%=pp;if(a<0)a+=pp;return a;}
28 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;}
29 void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
30 //void add(int x,int y,int z){ v[++e]=y; next[e]=head[x]; head[x]=e; cost[e]=z; }
31 int dx[5]={0,-1,1,0,0},dy[5]={0,0,0,-1,1};
32 ll read(){ ll ans=0; char last=' ',ch=getchar();
33 while(ch<'0' || ch>'9')last=ch,ch=getchar();
34 while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
35 if(last=='-')ans=-ans; return ans;
36 }
37 const int N=100005;
38 ll X[N],Y[N],A[N],B[N],sA[N],sB[N],ans=1e15;
39 int main(){
40     int n=read(),a,b;
41     for (int i=1;i<=n;i++){
42         a=read(),b=read();
43         X[i]=a+b; Y[i]=a-b;
44         A[i]=a+b; B[i]=a-b;
45     }
46     sort(A+1,A+n+1);
47     sort(B+1,B+n+1);
48     for (int i=1;i<=n;i++){
49         sA[i]=sA[i-1]+A[i]; sB[i]=sB[i-1]+B[i];
50     }
51     int p,q;
52     for (int i=1;i<=n;i++){
53         p=lower_bound(A+1,A+n+1,X[i])-A; q=lower_bound(B+1,B+n+1,Y[i])-B;
54         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)));
55     }
56     cout<<ans/2;
57     return 0;
58 }
View Code

 

posted @ 2017-10-17 18:43  SXia  阅读(164)  评论(0编辑  收藏  举报