POJ-3714 Raid 平面最近点对

　　分治算法修改该为两个点集的情况就可以了，加一个标记。。。
  1 //STATUS:C++_AC_2094MS_4880KB
  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=200010;
 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=1e30;
 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     double x,y;
 60     int id,index;
 61     Node(){}
 62     Node(double _x,double _y,int _index):x(_x),y(_y),index(_index){}
 63 }nod[N],temp[N];
 64 
 65 int n;
 66 
 67 double dist(Node &a,Node &b)
 68 {
 69     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 70 }
 71 
 72 bool cmpxy(Node a,Node b)
 73 {
 74     return a.x!=b.x?a.x<b.x:a.y<b.y;
 75 }
 76 
 77 bool 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     double 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     if(pn1.first==pn1.second || nod[pn1.first].index==nod[pn1.second].index)
 90         d1=OO;
 91     else d1=dist(nod[pn1.first],nod[pn1.second]);
 92     if(pn2.first==pn2.second || nod[pn2.first].index==nod[pn2.second].index)
 93         d2=OO;
 94     else d2=dist(nod[pn2.first],nod[pn2.second]);
 95 
 96     pii ret;
 97     d=Min(d1,d2);
 98     ret=d1<d2?pn1:pn2;
 99 
100     for(i=l,k=0;i<=r;i++){
101         if(fabs(nod[mid].x-nod[i].x)<=d){
102             temp[k++]=nod[i];
103         }
104     }
105     sort(temp,temp+k,cmpy);
106     for(i=0;i<k;i++){
107         for(j=i+1;j<k && fabs(temp[j].y-temp[i].y)<d;j++){
108             if(dist(temp[i],temp[j])<d && (temp[i].index^temp[j].index)){
109                 d=dist(temp[i],temp[j]);
110                 ret=make_pair(temp[i].id,temp[j].id);
111             }
112         }
113     }
114     return ret;
115 }
116 
117 void Init()
118 {
119     int i,t;
120     double x,y;
121     scanf("%d",&t);
122     n=t<<1;
123     for(i=0;i<t;i++){
124         scanf("%lf%lf",&x,&y);
125         nod[i]=Node(x,y,0);
126     }
127     for(;i<n;i++){
128         scanf("%lf%lf",&x,&y);
129         nod[i]=Node(x,y,1);
130     }
131     sort(nod,nod+n,cmpxy);
132     for(i=0;i<n;i++)nod[i].id=i;
133 }
134 
135 int main(){
136  //   freopen("in.txt","r",stdin);
137     int T,i,j;
138     scanf("%d",&T);
139     while(T--)
140     {
141         Init();
142         pii ans=Closest_Pair(0,n-1);
143 
144         printf("%.3lf
",dist(nod[ans.first],nod[ans.second]));
145     }
146     return 0;
147 }