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。。

 

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 using namespace std;
  7 #define  N 500005
  8 #define LL long long
  9 #define INF  1000000000000000000
 10 int mm;
 11 LL d;
 12 struct Point
 13 {
 14     LL x;
 15     LL y;
 16     int p;
 17 }point[N],tt[N];
 18 int tmpt[N];
 19 
 20 bool cmpxy(const Point& a, const Point& b)
 21 {
 22     if(a.x != b.x)
 23         return a.x < b.x;
 24     return a.y < b.y;
 25 }
 26 bool cmpy(const int& a, const int& b)
 27 {
 28     return point[a].y < point[b].y;
 29 }
 30 
 31 LL min(LL a, LL b)
 32 {
 33     return a < b ? a : b;
 34 }
 35 
 36 LL dis(int i, int j)
 37 {
 38     return (point[i].x-point[j].x)*(point[i].x-point[j].x)
 39                 + (point[i].y-point[j].y)*(point[i].y-point[j].y);
 40 }
 41 void Closest_Pair(int left, int right)
 42 {
 43     if(left==right)
 44     {
 45         return ;
 46     }
 47     if(left + 1 == right)
 48     {
 49          LL xx = dis(left, right);
 50          if(d>xx)
 51          {
 52 
 53              d = xx;
 54              mm = max(point[left].p,point[right].p);
 55          }
 56          return ;
 57      }
 58     int mid = (left+right)>>1;
 59     Closest_Pair(left,mid);
 60     Closest_Pair(mid+1,right);
 61     int i,j,k=0;
 62     for(i = left; i <= right; i++)
 63     {
 64         if((point[mid].x-point[i].x)*(point[mid].x-point[i].x) <= d)
 65             tmpt[k++] = i;
 66     }
 67     sort(tmpt,tmpt+k,cmpy);
 68     for(i = 0; i < k; i++)
 69     {
 70         for(j = i+1; j < k && (point[tmpt[j]].y-point[tmpt[i]].y)*(point[tmpt[j]].y-point[tmpt[i]].y)<d; j++)
 71         {
 72             LL d3 = dis(tmpt[i],tmpt[j]);
 73             if(d > d3)
 74             {
 75                  d = d3;
 76                  mm = max(point[tmpt[j]].p,point[tmpt[i]].p);
 77             }
 78         }
 79     }
 80 }
 81 int main()
 82 {
 83     int t,a[10],i;
 84     cin>>t;
 85     while(t--)
 86     {
 87         for(i = 1; i <= 7 ; i++)
 88         cin>>a[i];
 89         point[i].x = 0;
 90         point[i].y = 0;
 91         for(i = 1; i <= a[1] ; i++)
 92         {
 93             point[i].x = (point[i-1].x*a[2]+a[3])%a[4];
 94             point[i].y = (point[i-1].y*a[5]+a[6])%a[7];
 95             point[i].p = i;
 96             tt[i].x = point[i].x;
 97             tt[i].y = point[i].y;
 98             tt[i].p = i;
 99         }
100         LL s=0;
101         int n = a[1];
102         while(1)
103         {
104             sort(point+1,point+n+1,cmpxy);
105             d = INF;
106             Closest_Pair(1,n);
107             s+=(n-mm+1)*d;
108             n = mm-1;
109             for(i = 1 ; i <= n ; i++)
110             {
111                 point[i].x = tt[i].x;
112                 point[i].y = tt[i].y;
113                 point[i].p = tt[i].p;
114             }
115             if(n<=1)
116             break;
117         }
118         cout<<s<<endl;
119     }
120     return 0;
121 }
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posted @ 2013-07-31 19:23  _雨  阅读(188)  评论(0)  编辑  收藏  举报