Codeforces Round #319 (Div. 1) C. Points on Plane 分块-白红宇

Codeforces Round #319 (Div. 1) C. Points on Plane 分块

阅读量：6612 次

发布时间：2019-06-24

本文共 2265 字，大约阅读时间需要 7 分钟。

C. Points on Plane

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/576/problem/C

Description

On a plane are n points (xi, yi) with integer coordinates between 0 and 106. The distance between the two points with numbers a and bis said to be the following value: $\text{[math]}$ (the distance calculated by such formula is called Manhattan distance).

We call a hamiltonian path to be some permutation pi of numbers from 1 to n. We say that the length of this path is value $\text{[math]}$ .

Find some hamiltonian path with a length of no more than 25 × 108. Note that you do not have to minimize the path length.

Input

The first line contains integer n (1 ≤ n ≤ 106).

The i + 1-th line contains the coordinates of the i-th point: xi and yi (0 ≤ xi, yi ≤ 106).

It is guaranteed that no two points coincide.

Output

Print the permutation of numbers pi from 1 to n — the sought Hamiltonian path. The permutation must meet the inequality $\text{[math]}$ .

If there are multiple possible answers, print any of them.

It is guaranteed that the answer exists.

Sample Input

0 7

8 10

3 4

5 0

9 12

Sample Output

4 3 1 2 5

HINT

In the sample test the total distance is:

$\text{[math]}$

(|5 - 3| + |0 - 4|) + (|3 - 0| + |4 - 7|) + (|0 - 8| + |7 - 10|) + (|8 - 9| + |10 - 12|) = 2 + 4 + 3 + 3 + 8 + 3 + 1 + 2 = 26

题意

给你一个曼哈顿距离的图，然后要求你找到一个链，链穿了所有的点

然后要求这链的长度<=25*10e8

题解：

就分块咯，分成1000块，每个块内y坐标最多走10e6长度，x坐标最多走n*10e3个，n表示一块内的点数

n是一个二次函数维护的东西……所以大概答案最后就是10e3(10e6+10e6) = 2*10e9

所以大概看看脸，就能把这道题AC了

代码：

//qscqesze#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  #include 
                  #include 
                   
                    typedef long long ll;using namespace std;//freopen("D.in","r",stdin);//freopen("D.out","w",stdout);#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)#define maxn 1000500#define mod 1001#define eps 1e-9#define pi 3.1415926int Num;//const int inf=0x7fffffff;const ll inf=999999999;inline ll read(){ ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){ if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f;}//*************************************************************************************pair
                    
                     
                       > p[maxn];int main(){ int n=read(); for(int i=0;i

转载地址：http://tzaso.baihongyu.com/

你可能感兴趣的文章