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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