E. Points on Plane
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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 b is said to be the following value: 
(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 
.
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 
.
If there are multiple possible answers, print any of them.
It is guaranteed that the answer exists.
Examples
input
5
0 7
8 10
3 4
5 0
9 12
output
4 3 1 2 5
Note
In the sample test the total distance is:
(|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：
给你二维平面上的n个坐标，要你经过所有的点，但是总路程小于25*1e8。（按顺序的两个坐标的路程就是两坐标的距离）
输出按下标标号的经过顺序。
做法：
一开始看到题面可能会有点懵，只要总路程小于25*1e8就好，但是这道题有100w个点，点的坐标范围也是在（0，0）到（1e6,1e6）的，所以如果样例比较极端，再随便连的话超过25*1e8是分分钟的事情，所以要进行优化。
因为连点肯定是离得越近越好，一个区域连过之后就不要再回来了，按照类似从左到右的顺序连边。因此就用到了莫队的分块思想，这里有个莫队的优化，如果分到的块是奇数块，那y坐标从小到大，偶数块从大到小，这样路线就会变成
可以达到我们的要求。
代码很短如下：
#include<bits/stdc++.h>
using namespace std;
struct node{
int x,y,lev,id;
bool operator < (const node &a) const {
if(lev==a.lev) {
if(lev%2) return y<a.y;
return y>a.y;
}
return lev<a.lev;
}
}e[1000005];
int n,biao;
int main(){
scanf("%d",&n);
biao=sqrt(n);
for(int i=1;i<=n;i++){
scanf("%d%d",&e[i].x,&e[i].y);
e[i].id=i,e[i].lev=e[i].x/biao;
}
sort(e+1,e+1+n);
for(int i=1;i<=n;i++){
printf("%d%c",e[i].id,i==n?'\n':' ');
}
return 0;
}