hdu 1007_平面最近点对模板
简单题目,直接模板上
http://acm.hdu.edu.cn/showproblem.php?pid=1007 Quoit Design
O(n * log(n) *log(n) )的复杂度
[cpp] view plaincopy
#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
#include <bitset>
#include <fstream>
using namespace std;
//LOOP
#define FF(i, a, b) for(int i = (a); i < (b); ++i)
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
//OTHER
#define SZ(V) (int)V.size()
#define PB push_back
#define MP make_pair
#define all(x) (x).begin(),(x).end()
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RS(s) scanf("%s", s)
//OUTPUT
#define WI(n) printf("%d\n", n)
#define WS(n) printf("%s\n", n)
//debug
//#define online_judge
#ifndef online_judge
#define debugt(a) cout << (#a) << "=" << a << " ";
#define debugI(a) debugt(a) cout << endl
#define debugII(a, b) debugt(a) debugt(b) cout << endl
#define debugIII(a, b, c) debugt(a) debugt(b) debugt(c) cout << endl
#define debugIV(a, b, c, d) debugt(a) debugt(b) debugt(c) debugt(d) cout << endl
#else
#define debugI(v)
#define debugII(a, b)
#define debugIII(a, b, c)
#define debugIV(a, b, c, d)
#endif
#define sqr(x) (x) * (x)
typedef long long LL;
typedef unsigned long long ULL;
typedef vector <int> VI;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-10;
const int MOD = 100000007;
const int MAXN = 100010;
const double PI = acos(-1.0);
inline int dcmp(double x)
{
if(fabs(x) < EPS) return 0;
else return x < 0 ? -1 : 1;
}
struct Point
{
double x, y;
Point(double x=0, double y=0):x(x),y(y) { }
inline void read()
{
scanf("%lf%lf", &x, &y);
}
};
//最近点对
Point point[MAXN];
int tmpt[MAXN], Y[MAXN];
inline bool cmpxy(Point a, Point b)
{
if(a.x != b.x)
return a.x < b.x;
return a.y < b.y;
}
inline bool cmpy(int a, int b)
{
return point[a].y < point[b].y;
}
inline double dist(int x, int y)
{
Point& a = point[x], &b = point[y];
return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));
}
double Closest_Pair(int left, int right)
{
double d = 1e100;
if(left == right)
return d;
if(left + 1 == right)
return dist(left, right);
int mid = (left + right) >> 1;
double d1 = Closest_Pair(left, mid);
double d2 = Closest_Pair(mid + 1, right);
d = min(d1, d2);
int k = 0;
//分离出宽度为d的区间
FE(i, left, right)
{
if(fabs(point[mid].x - point[i].x) <= d)
tmpt[k++] = i;
}
sort(tmpt, tmpt + k, cmpy);
//线性扫描
REP(i, k)
{
for(int j = i + 1; j < k && point[tmpt[j]].y-point[tmpt[i]].y < d; j++)
{
double d3 = dist(tmpt[i],tmpt[j]);
if(d > d3)
d = d3;
}
}
return d;
}
int main()
{
//freopen("input.txt", "r", stdin);
int n;
while (~RI(n) && n)
{
REP(i, n)
{
point[i].read();
}
sort(point, point + n, cmpxy);
printf("%.2f\n", Closest_Pair(0, n - 1) / 2);
}
return 0;
}
补充：软件开发 , C++ ,
