POJ-3714 Raid 平面最近点对
题目链接：http://poj.org/problem?id=3714
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\n",dist(nod[ans.first],nod[ans.second])); 145 } 146 return 0; 147 }
