HDU4631 Sad Love Story（最近点对）转自网络

转自http://blog.csdn.net/acm_zl/article/details/9634443

这道题目的意思就是不断加入n个点。当点数>=2的时候，每加入一个点累加两点间最近距离的平方。按照给定的Ax,Bx,Cx,Ay,By,Cy,以及递推式可以产生n个点。
所有首先n个点，做一下最近点对，复杂度O(nlogn)然后产生的最近点对，对于编号在最近点对后面的结果都可以累加了，同时后面的点也不需要了。所有去掉一部分点再次进行最近点对。这样不断重复，直到剩下一个点为止。

#include <stdio.h>  #include <algorithm>  #include <string.h>  #include <iostream>  #include <map>  #include <vector>  #include <queue>  #include <set>  #include <string>  #include <math.h>  using namespace std;  const int MAXN = 500010;  struct Point  {      int x,y;      int id;      int index;      Point(int _x = 0,int _y = 0,int _index = 0)      {          x = _x;          y = _y;          index = _index;      }  };  Point P[MAXN];  long long dist(Point a,Point b)  {      return ((long long)(a.x-b.x)*(a.x-b.x) + (long long)(a.y-b.y)*(a.y-b.y));  }  Point p[MAXN];  Point tmpt[MAXN];  bool cmpxy(Point a,Point b)  {      if(a.x != b.x)return a.x < b.x;      else return a.y < b.y;  }  bool cmpy(Point a,Point b)  {      return a.y < b.y;  }  pair<int,int> Closest_Pair(int left,int right)  {      long long d = (1LL<<50);      if(left == right)return make_pair(left,right);      if(left + 1 == right)          return make_pair(left,right);      int mid = (left+right)/2;      pair<int,int>pr1 = Closest_Pair(left,mid);      pair<int,int>pr2 = Closest_Pair(mid+1,right);      long long d1,d2;      if(pr1.first == pr1.second)          d1 = (1LL<<50);      else d1 = dist(p[pr1.first],p[pr1.second]);      if(pr2.first == pr2.second)          d2 = (1LL<<50);      else d2 = dist(p[pr2.first],p[pr2.second]);      pair<int,int>ans;      if(d1 < d2)ans = pr1;      else ans = pr2;      d = min(d1,d2);      int k = 0;      for(int i = left;i <= right;i++)      {          if((long long)(p[mid].x - p[i].x)*(p[mid].x - p[i].x) <= d)              tmpt[k++] = p[i];      }      sort(tmpt,tmpt+k,cmpy);      for(int i = 0;i <k;i++)      {          for(int j = i+1;j < k && (long long)(tmpt[j].y - tmpt[i].y)*(tmpt[j].y - tmpt[i].y) < d;j++)          {              if(d > dist(tmpt[i],tmpt[j]))              {                  d = dist(tmpt[i],tmpt[j]);                  ans = make_pair(tmpt[i].id,tmpt[j].id);              }          }      }      return ans;  }  int main()  {       int T;      int n,Ax,Ay,Bx,By,Cx,Cy;      scanf("%d",&T);      while(T--)      {          scanf("%d%d%d%d%d%d%d",&n,&Ax,&Bx,&Cx,&Ay,&By,&Cy);          P[0] = Point(0,0,0);          for(int i = 1;i <= n;i++)          {              long long x= ((long long)P[i-1].x*Ax + Bx)%Cx;              long long y = ((long long)P[i-1].y*Ay + By)%Cy;              P[i] = Point(x,y,i);          }          int end = n;          long long ans = 0;          while(end > 1)          {              for(int i = 0;i < end;i++)                  p[i] = P[i+1];              sort(p,p+end,cmpxy);              for(int i = 0;i < end;i++)                  p[i].id = i;              pair<int,int>pr = Closest_Pair(0,end-1);              int Max = max(p[pr.first].index,p[pr.second].index);              ans += (long long)(end-Max+1)*dist(p[pr.first],p[pr.second]);              end = Max-1;          }          printf("%I64d\n",ans);      }      return 0;  }