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AcWing - 119 - 袭击 = 平面最近点对

https://www.acwing.com/problem/content/121/

给两种点,黑点和红点各n(n<=1e5)个,求最近的黑点和红点之间的距离,xy在1e9范围内。

随机旋转法。先sort一遍更新d使得d尽可能变小,然后randomshuffle(实际上貌似只需要randomshuffle其中一个就够了),再更新一边d使得d尽可能变小。

然后随机乱旋转一通,根据已经缩小的d找出可能更新答案的上下界(二分找出来),然后暴力更新,随着更新的过程进行d可能会进一步变小,复杂度不能保证但是感觉问题应该不会很大。何况是randomshuffle之后又随机乱转一通,什么精心构造的数据应该都完蛋了。

#include<bits/stdc++.h>
using namespace std;

const double PI = acos(-1.0);

struct Point {
    double x, y;
    Point() {}
    inline bool operator<(const Point &b) const {
        return x < b.x;
    }
} p[100000 + 5], q[100000 + 5], tmp;

int n;

inline double RandomDouble() {
    return 1.0 * rand() / RAND_MAX;
}

inline bool cmp(const Point &a, const Point &b) {
    return a.x < b.x;
}

inline double dis(const Point &a, const Point &b) {
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

double Calc(const double &A, double d) {
    double x0 = -1e9 + 2e9 * RandomDouble(), y0 = -1e9 + 2e9 * RandomDouble(); //随机弄一个旋转原点
    double cosA = cos(A), sinA = sin(A);
    double xc = -x0 * cosA + y0 * sinA + x0;
    double yc = -x0 * sinA - y0 * cosA + y0;
    //利用图形学的知识加速
    for(int i = 1; i <= n; ++i) {
        double x = p[i].x, y = p[i].y;
        p[i].x = x * cosA - y * sinA + xc;
        p[i].y = x * sinA + y * cosA + yc;
    }
    sort(p + 1, p + 1 + n);
    for(int i = 1; i <= n; i++) {
        double x = q[i].x, y = q[i].y;
        q[i].x = x * cosA - y * sinA + xc;
        q[i].y = x * sinA + y * cosA + yc;
    }
    sort(q + 1, q + 1 + n);

    for(int i = 1; i <= n; ++i) {
        tmp.x = p[i].x - d;
        int minj = max(1, int(lower_bound(q + 1, q + 1 + n, tmp) - q));
        tmp.x = p[i].x + d;
        int maxj = min(n, int(lower_bound(q + 1, q + 1 + n, tmp) - q));
        for(int j = minj; j <= maxj && q[j].x - p[i].x < d; j++)
            d = min(d, dis(p[i], q[j]));
    }
    return d;
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    srand(time(0));
    int t;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i)
            scanf("%lf%lf", &p[i].x, &p[i].y);
        for(int i = 1; i <= n; ++i)
            scanf("%lf%lf", &q[i].x, &q[i].y);

        double d = 1e36;
        sort(p + 1, p + 1 + n);
        sort(q + 1, q + 1 + n);
        for(int i = 1; i <= n; ++i) {
            int c = min(n, i + 5);
            for(int j = max(1, i - 5); j <= c; ++j) {
                d = min(d, dis(p[i], q[j]));
            }
        }
        random_shuffle(p + 1, p + 1 + n);
        random_shuffle(q + 1, q + 1 + n);
        for(int i = 1; i <= n; ++i) {
            int c = min(n, i + 5);
            for(int j = max(1, i - 5); j <= c; ++j) {
                d = min(d, dis(p[i], q[j]));
            }
        }
        d = Calc(RandomDouble() * 2.0 * PI, d);
        //d = Calc(RandomDouble() * 2.0 * PI, d);
        //d = Calc(RandomDouble() * 2.0 * PI, d);
        printf("%.3f\n", d);
    }
    return 0;
}
posted @ 2019-09-13 10:50 Inko 阅读(...) 评论(...) 编辑 收藏