zoukankan      html  css  js  c++  java
  • 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 }
  • 相关阅读:
    大牛思考方式
    web面试题大全
    github上最全的资源教程-前端涉及的所有知识体系
    java switch语句注意的事项
    Lucene Payload 的研究与应用
    hive array、map、struct使用
    黑马程序员--正则表达式
    [置顶] 读源码练内功(一):guava之eventbus
    自定义Java Annotations实例以及用Java Reflection来解析自定义的Annotation
    Solr之NamedList 简单介绍与实例解析
  • 原文地址:https://www.cnblogs.com/zhsl/p/3234692.html
Copyright © 2011-2022 走看看