hdu4631Sad Love Story(多校3）（最接近点对）

http://acm.hdu.edu.cn/showproblem.php?pid=4631

比赛的时候搜到了最接近点对的求法 Nlog(N） 又估摸着依次插入求的话会TLE 想了想觉得可以先把最近的位置求出来 然后后面的直接不用求了 依次直到减完 又觉得可能会有变态的数据每次最近的都在最后面 没敢写。。后来 发现它出现在题解的方法三中。。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 #define  N 500005
 #define LL long long
 #define INF  1000000000000000000
 int mm;
 LL d;
 struct Point
 {
     LL x;
     LL y;
     int p;
 }point[N],tt[N];
 int tmpt[N];

 bool cmpxy(const Point& a, const Point& b)
 {
     if(a.x != b.x)
         return a.x < b.x;
     return a.y < b.y;
 }
 bool cmpy(const int& a, const int& b)
 {
     return point[a].y < point[b].y;
 }

 LL min(LL a, LL b)
 {
     return a < b ? a : b;
 }

 LL dis(int i, int j)
 {
     return (point[i].x-point[j].x)*(point[i].x-point[j].x)
                 + (point[i].y-point[j].y)*(point[i].y-point[j].y);
 }
 void Closest_Pair(int left, int right)
 {
     if(left==right)
     {
         return ;
     }
     if(left +  == right)
     {
          LL xx = dis(left, right);
          if(d>xx)
          {

              d = xx;
              mm = max(point[left].p,point[right].p);
          }
          return ;
      }
     int mid = (left+right)>>;
     Closest_Pair(left,mid);
     Closest_Pair(mid+,right);
     int i,j,k=;
     for(i = left; i <= right; i++)
     {
         if((point[mid].x-point[i].x)*(point[mid].x-point[i].x) <= d)
             tmpt[k++] = i;
     }
     sort(tmpt,tmpt+k,cmpy);
     for(i = ; i < k; i++)
     {
         for(j = i+; j < k && (point[tmpt[j]].y-point[tmpt[i]].y)*(point[tmpt[j]].y-point[tmpt[i]].y)<d; j++)
         {
             LL d3 = dis(tmpt[i],tmpt[j]);
             if(d > d3)
             {
                  d = d3;
                  mm = max(point[tmpt[j]].p,point[tmpt[i]].p);
             }
         }
     }
 }
 int main()
 {
     int t,a[],i;
     cin>>t;
     while(t--)
     {
         for(i = ; i <=  ; i++)
         cin>>a[i];
         point[i].x = ;
         point[i].y = ;
         for(i = ; i <= a[] ; i++)
         {
             point[i].x = (point[i-].x*a[]+a[])%a[];
             point[i].y = (point[i-].y*a[]+a[])%a[];
             point[i].p = i;
             tt[i].x = point[i].x;
             tt[i].y = point[i].y;
             tt[i].p = i;
         }
         LL s=;
         int n = a[];
         while()
         {
             sort(point+,point+n+,cmpxy);
             d = INF;
             Closest_Pair(,n);
             s+=(n-mm+)*d;
             n = mm-;
             for(i =  ; i <= n ; i++)
             {
                 point[i].x = tt[i].x;
                 point[i].y = tt[i].y;
                 point[i].p = tt[i].p;
             }
             if(n<=)
             break;
         }
         cout<<s<<endl;
     }
     return ;
 }