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
",dist(nod[ans.first],nod[ans.second])/2);
136     }
137     return 0;
138 }