【CF1237C】Balanced Removals（降维）

题意：三维平面上有n个点，每个点的坐标为（x[i]，y[i]，z[i]），n为偶数

现在要求取n/2次，每次取走一对点（x,y），要求没有未被取走的点在以x和y为对角点的矩形中

要求给出任意一组合法方案

n<=5e4，abs(x[i],y[i],z[i])<=1e8

思路：我觉得托老爷的官方题解的google机翻已经够简明了

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 //typedef pair<ll,ll>P;
 #define N  200010
 #define M  200010
 #define fi first
 #define se second
 #define MP make_pair
 #define pb push_back
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1

 const ll MOD=1e9+,inv2=(MOD+)/;
       double eps=1e-;
       ll INF=1e15;
       int dx[]={-,,,};
       int dy[]={,,-,};

 struct node
 {
     int x,y,z,id;
 }a[N],b[N];

 bool cmp(node a,node b)
 {
     if(a.x!=b.x) return a.x<b.x;
     if(a.y!=b.y) return a.y<b.y;
     return a.z<b.z;
 }

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 int main()
 {
     int n=read();
     rep(i,,n)
     {
         a[i].x=read(),a[i].y=read(),a[i].z=read();
         a[i].id=i;
     }
     sort(a+,a+n+,cmp);
     int i,j,k,m=;
     for(i=;i<=n;)
     {
         j=i;
         while(j<=n&&a[i].x==a[j].x&&a[i].y==a[j].y) j++;
         for(k=i;k+<=j;k+=) printf("%d %d\n",a[k].id,a[k+].id);
         if(k==j-) b[++m]=a[k];
         i=j;
     }
     n=;
     for(i=;i<=m;)
     {
         j=i;
         while(j<=m&&b[i].x==b[j].x) j++;
         for(k=i;k+<=j;k+=) printf("%d %d\n",b[k].id,b[k+].id);
         if(k==j-) a[++n]=b[k];
         i=j;
     }
     for(i=;i<=n;i+=) printf("%d %d\n",a[i].id,a[i+].id);
     return ;
 }