HDU-1007 Quoit Design 平面最近点对
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1007
简单裸题，测测模板，G++速度快了不少，应该是编译的时候对比C++优化了不少。。
1 //STATUS:G++_AC_1703MS_5004KB 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=100010; 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 struct Node{ //id为nod排序后的编号值，index为排序前的标号值(随便自己定义) 58 double x,y; 59 int id,index; 60 Node(){} 61 Node(double _x,double _y,int _index):x(_x),y(_y),index(_index){} 62 }nod[N],temp[N]; 63 64 int n; 65 66 double dist(Node &a,Node &b) 67 { 68 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 69 } 70 71 bool cmpxy(Node a,Node b) //先按x排序,然后按y排序 72 { 73 return a.x!=b.x?a.x<b.x:a.y<b.y; 74 } 75 76 bool cmpy(Node a,Node b) //按y排序 77 { 78 return a.y<b.y; 79 } 80 81 pii Closest_Pair(int l,int r) //返回排序后点的编号 82 { 83 if(l==r || l+1==r)return pii(l,r); //只有一个点或者两个点 84 double d,d1,d2; 85 int i,j,k,mid=(l+r)/2; 86 //左右两边最小距离点的编号 87 pii pn1=Closest_Pair(l,mid); 88 pii pn2=Closest_Pair(mid+1,r); 89 //左右两遍的最小距离 90 d1=(pn1.first==pn1.second?OO:dist(nod[pn1.first],nod[pn1.second])); 91 d2=(pn2.first==pn2.second?OO:dist(nod[pn2.first],nod[pn2.second])); 92 pii ret; 93 d=Min(d1,d2); 94 ret=d1<d2?pn1:pn2; 95 //找出在mid-d,mid+d之间的点 96 for(i=l,k=0;i<=r;i++){ 97 if(fabs(nod[mid].x-nod[i].x)<=d){ 98 temp[k++]=nod[i]; 99 } 100 } 101 //合并两边,求最小距离 102 sort(temp,temp+k,cmpy); 103 for(i=0;i<k;i++){ 104 for(j=i+1;j<k && fabs(temp[j].y-temp[i].y)<d;j++){ 105 if(dist(temp[i],temp[j])<d){ 106 d=dist(temp[i],temp[j]); 107 ret=make_pair(temp[i].id,temp[j].id); 108 } 109 } 110 } 111 112 return ret; 113 } 114 115 void Init() //初始化点 116 { 117 int i; 118 double x,y; 119 for(i=0;i<n;i++){ 120 scanf("%lf%lf",&x,&y); 121 nod[i]=Node(x,y,i); 122 } 123 sort(nod,nod+n,cmpxy); 124 for(i=0;i<n;i++)nod[i].id=i; //排序后节点编号 125 } 126 127 int main(){ 128 // freopen("in.txt","r",stdin); 129 int T,i,j; 130 while(~scanf("%d",&n) && n) 131 { 132 Init(); 133 pii ans=Closest_Pair(0,n-1); 134 135 printf("%.2lf\n",dist(nod[ans.first],nod[ans.second])/2); 136 } 137 return 0; 138 }
