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

复杂度O( n * log(n) )

#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);    }};Point p[MAXN];int px[MAXN], py[MAXN], ty[MAXN];inline int cmpx(Point a, Point b){    return a.x < b.x;}inline int cmpy(int a, int b){    return p[a].y < p[b].y;}inline double min(double a, double b){    return a < b ? a : b;}inline double dist2(Point a, Point b){    return sqr(a.x - b.x) + sqr(a.y - b.y);}double min_dis = 1e100;//返回最小距离的平方double mindist(int* X, int* Y, int size){    if(size <= 3)    {        if(size == 2)            return dist2(p[X[0]], p[X[1]]);        REP(i, size)            min_dis = min(min_dis, dist2(p[X[i]], p[X[(i + 1) % size]]));        return min_dis;    }    int pr = size >> 1, pl = size - pr;    int l1 = 0, l2 = pl;    REP(i, size)        if(Y[i] < X[pl])            ty[l1++] = Y[i];        else            ty[l2++] = Y[i];    REP(i, size)        Y[i] = ty[i];    min_dis = min(mindist(X, Y, pl), mindist(X + pl, Y + pl, pr));    l1 = 0;    REP(i, size)        if(sqr(p[Y[i]].x - p[X[pl - 1]].x) <= min_dis)            ty[l1++] = Y[i];    REP(i, l1)        for(int j = 1; i + j < l1 && sqr(p[ty[i + j]].y - p[ty[i]].y) <= min_dis; j++)            min_dis = min(min_dis, dist2(p[ty[i]], p[ty[i + j]]));    return min_dis;}//px的初始化//每次调用时,两次排序,mid_dis赋值int main(){    int n;    REP(i, MAXN) px[i] = i;    while(scanf("%d",&n)&&n)    {        REP(i, n)        {            p[i].read();            py[i] = i;        }        sort(p, p + n, cmpx);        sort(py, py + n, cmpy);        min_dis = 1e100;        printf("%.2f\n", sqrt(mindist(px, py, n)) / 2);    }    return 0;}

O(n * log(n) *log(n) )的复杂度

#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;}