CodeForces 576C Points on Plane 莫队

CodeForces 576C Points on Plane

给你n个点,把这些点按一定顺序排起来,使得最后相邻两个点的曼哈顿距离之和小于2.5*10^9.
输出这些点的排列顺序.

/*
曼哈顿距离最小生成树.明显了这就是个莫队思路题.
我们把x坐标看做询问左端点,y坐标看做询问右端点,对点进行分块.
排序的时候按x坐标所在块按从小到大排,相等按y坐标从小到大排,然后就被叉掉了.
*/
#include<bits/stdc++.h> //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define rec register char
#define rel register ll
#define gc getchar
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
inline int read(){
  int x=0,f=1;char c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return f?x:-x;
  }
template <typename mitsuha>
inline bool read(mitsuha &x){
  x=0;int f=1;char c=gc();
  for (;!isdigit(c)&&~c;c=gc()) f^=c=='-';
  if (!~c) return 0;
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return x=f?x:-x,1;
  }
template <typename mitsuha>
inline int write(mitsuha x){
  if (!x) return 0&pc(48);
  if (x<0) x=-x,pc('-');
  int bit[20],i,p=0;
  for (;x;x/=10) bit[++p]=x%10;
  for (i=p;i;--i) pc(bit[i]+48);
  return 0;
  }
inline char fuhao(){
  char c=gc();
  for (;isspace(c);c=gc());
  return c;
  }
}using namespace chtholly;
using namespace std;
const int yuzu=1e6,block=1e3;
#define bl(x) (((x)-1)/block+1) 
#define ls(x) (((x)-1)*block+1)
#define rs(x) (min(n,(x)*block))

struct point{
int x,y,id;
void rd(int i){x=read(),y=read(),id=i;}
bool operator <(const point &b) const{
  return bl(x)^bl(b.x)?x<b.x:y<b.y;
  }
}a[yuzu|10];

int main(){
int i,n=read();
for (i=1;i<=n;++i) a[i].rd(i);
int num=n/block;
sort(a+1,a+n+1);//一个sort世界清静.
for (i=1;i<=n;++i) write(a[i].id),p32;  
}

谢谢大家.