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C. Points on Plane
题目连接
http://codeforces.com/contest/576/problem/CDescription
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: 
(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.
Sample Input
5
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:
(|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：给定坐标系中n个点的坐标（范围[0,10^6]），求一种连边形成链后总长度<=2.5*10^9 的方案。n<=10^6。链的长度是链上相邻两点的曼哈顿距离和。（输出任意解即可）
题解：将X轴的点分成1000块，这样每块内Y坐标可走10^6,X坐标可走10^3,每块之间的连接最长为2*10^3，所以最大总距离为2*10^9+2*10^6，符合要求。
AC代码：
#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
inline int read()
{int 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;
}
int n;
const int maxn = 1000007;
pair<int,pair<int,int> > Q[maxn];
int main()
{n = read();for(int i=1;i<=n;i++){Q[i].first = read()/1050;Q[i].second.first = read();Q[i].second.second = i;}sort(Q+1,Q+1+n);for(int i=1;i<=n;i++){cout<<Q[i].second.second<<" ";}cout<<endl;return 0;
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