HDU-4631 Sad Love Story 平面最近点对

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4631
　　数据是随机的，没有极端数据，所以可以分段考虑，最小值是一个单调不增的函数，然后每次分治算平面最近点对就可以了。。。
  1 //STATUS:G++_AC_10390MS_23804KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //#pragma comment(linker,"/STACK:102400000,102400000")
 25 //using namespace __gnu_cxx;
 26 //define
 27 #define pii pair<int,int>
 28 #define mem(a,b) memset(a,b,sizeof(a))
 29 #define lson l,mid,rt<<1
 30 #define rson mid+1,r,rt<<1|1
 31 #define PI acos(-1.0)
 32 //typedef
 33 typedef __int64 LL;
 34 typedef unsigned __int64 ULL;
 35 //const
 36 const int N=500010;
 37 const int INF=0x3f3f3f3f;
 38 const int MOD=10007,STA=8000010;
 39 const LL LNF=1LL<<55;
 40 const double EPS=1e-8;
 41 const double OO=1e15;
 42 const int dx[4]={-1,0,1,0};
 43 const int dy[4]={0,1,0,-1};
 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 45 //Daily Use ...
 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 56 //End
 57 
 58 struct Node{
 59     LL x,y;
 60     LL id,index;
 61     Node(){}
 62     Node(LL _x,LL _y,LL _index):x(_x),y(_y),index(_index){}
 63 }p[N],nod[N],temp[N];
 64 
 65 int n;
 66 
 67 LL dist(Node &a,Node &b)
 68 {
 69     return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
 70 }
 71 
 72 int cmpxy(Node a,Node b)
 73 {
 74     return a.x!=b.x?a.x<b.x:a.y<b.y;
 75 }
 76 
 77 int cmpy(Node a,Node b)
 78 {
 79     return a.y<b.y;
 80 }
 81 
 82 pii Closest_Pair(int l,int r)
 83 {
 84     if(l==r || l+1==r)return pii(l,r);
 85     LL d,d1,d2;
 86     int i,j,k,mid=(l+r)/2;
 87     pii pn1=Closest_Pair(l,mid);
 88     pii pn2=Closest_Pair(mid+1,r);
 89     d1=(pn1.first==pn1.second?LNF:dist(nod[pn1.first],nod[pn1.second]));
 90     d2=(pn2.first==pn2.second?LNF:dist(nod[pn2.first],nod[pn2.second]));
 91     pii ret;
 92     d=Min(d1,d2);
 93     ret=d1<d2?pn1:pn2;
 94     for(i=l,k=0;i<=r;i++){
 95         if((nod[mid].x-nod[i].x)*(nod[mid].x-nod[i].x)<=d){
 96             temp[k++]=nod[i];
 97         }
 98     }
 99     sort(temp,temp+k,cmpy);
100     for(i=0;i<k;i++){
101         for(j=i+1;j<k && (temp[j].y-temp[i].y)*(temp[j].y-temp[i].y)<d;j++){
102             if(dist(temp[i],temp[j])<d){
103                 d=dist(temp[i],temp[j]);
104                 ret=make_pair(temp[i].id,temp[j].id);
105             }
106         }
107     }
108 
109     return ret;
110 }
111 
112 void Init()
113 {
114     int i;
115     LL x,y,Ax,Bx,Cx,Ay,By,Cy;
116     cin>>n>>Ax>>Bx>>Cx>>Ay>>By>>Cy;
117     x=y=0;
118     for(i=0;i<n;i++){
119         x=(x*Ax+Bx)%Cx;
120         y=(y*Ay+By)%Cy;
121         p[i]=Node(x,y,i);
122     }
123 }
124 
125 int main(){
126  //   freopen("in.txt","r",stdin);
127     int T,i,j,k;
128     LL ans,hig;
129     scanf("%d",&T);
130     while(T--)
131     {
132         Init();
133 
134         int end=n;
135         pii t;
136         ans=0;
137         while(end>0){
138             for(i=0;i<end;i++)nod[i]=p[i];
139             sort(nod,nod+end,cmpxy);
140             for(i=0;i<end;i++)nod[i].id=i;
141             t=Closest_Pair(0,end-1);
142             hig=Max(nod[t.first].index,nod[t.second].index);
143             ans+=(end-hig)*dist(nod[t.first],nod[t.second]);
144             end=hig;
145         }
146         cout<<ans<<endl;
147     }
148     return 0;
149 }