ZOJ 2107 HDU 1007 Quoit Design(最近點對)

最近點對的裸題

利用分治去搞搞即可

代碼：

#include 
#include 
#include 
#include 
using namespace std;

const int N = 100005;

struct Point {
    double x, y;
    void read() {
        scanf("%lf%lf", &x, &y);
    }
};

bool cmpx(Point a, Point b) {
    return a.x < b.x;
}

bool cmpy(Point a, Point b) {
    return a.y < b.y;
}

double dis(Point a, Point b) {
    double dx = a.x - b.x;
    double dy = a.y - b.y;
    return sqrt(dx * dx + dy * dy);
}

int n;

Point p[N], tmp[N];
int tn;

double gao(double ans) {
    sort(tmp, tmp + tn, cmpy);
    for (int i = 0; i < tn; i++) {
        for(int j = i + 1; j < tn && tmp[j].y - tmp[i].y < ans; j++) {
            ans = min(ans, dis(tmp[i], tmp[j]));
        }
    }
    return ans;
}

double ClosePoint(int l, int r) {
    if (r - l + 1 <= 3) {
        tn = 0;
        for (int i = l; i <= r; i++)
            tmp[tn++] = p[i];
        return gao(1e20);
    }
    int mid = (l + r)>>1;
    double s = min(ClosePoint(l, mid), ClosePoint(mid + 1, r));
    tn = 0;
    for (int i = l; i <= r; i++) {
        if (fabs(p[i].x - p[mid].x) < s)
            tmp[tn++] = p[i];
    }
    return gao(s);
}

int main() {
    while (~scanf("%d", &n) && n) {
        for (int i = 0; i < n; i++)
            p[i].read();
        sort(p, p + n, cmpx);
        printf("%.2f\n", ClosePoint(0, n - 1) / 2);
    }
    return 0;
}