模板 - 平面最近点对

Posted 2022-10-23 inko

这个是正确的算法，但是复杂度可能会卡到 \(O(n^2)\) ，加上每个点最多匹配的临近点最多15个/30个限制的话复杂度就可以保证了，最多就再做一次增加正确的几率，我确实不行从头到尾都是随机的怎么有人卡得掉。

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

const double PI = acos(-1.0);

struct Point 
    double x, y;
 p[200000 + 5];

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 - x0) * cosA - (y - y0) * sinA + x0;
        //p[i].y = (x - x0) * sinA + (y - y0) * cosA + y0;
        p[i].x = x * cosA - y * sinA + xc;
        p[i].y = x * sinA + y * cosA + yc;
    
    sort(p + 1, p + 1 + n, cmp);

    for(int i = 1; i <= n; i++) 
        for(int j = i + 1; j <= n /*&& j <= i + 15*/ && p[j].x - p[i].x < d; j++)
            d = min(d, dis(p[i], p[j]));
    
    return d;


int main() 
    srand(time(0));

    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
        scanf("%lf%lf", &p[i].x, &p[i].y);

    double d = 1e36;
    d = Calc(RandomDouble() * 2.0 * PI, d);
    //d = Calc(RandomDouble() * 2.0 * PI, d);
    printf("%.4f\n", d);
    return 0;

分治算法，虽然每一层里面都排序，但实际上选出的点并不多。

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

const double INF = 1e36;

const int MAXN = 200000;

int n, tmp[MAXN];
struct Point 
    double x, y;
 S[MAXN + 5];

bool cmpx(const Point &a, const Point&b) 
    return a.x < b.x;


bool cmpidy(const int &a, const int &b) 
    return S[a].y < S[b].y ;


inline double min(const double &a, const double &b) 
    return a < b ? a : b;


inline double dist(const int &i, const int &j) 
    return sqrt((S[i].x - S[j].x) * (S[i].x - S[j].x) + (S[i].y - S[j].y) * (S[i].y - S[j].y));


double merge(int left, int right) 
    double d = INF;
    if(left == right)
        return d ;
    if(left + 1 == right)
        return dist(left, right);
    int mid = left + right >> 1;
    double d1 = merge(left, mid) ;
    double d2 = merge(mid + 1, right) ;
    d = min(d1, d2);
    int i, j, k = 0;
    for(i = left; i <= right; i++)
        if(fabs(S[mid].x - S[i].x) <= d)
            tmp[++k] = i;
    sort(tmp + 1, tmp + k + 1, cmpidy);
    for(i = 1; i <= k; i++)
        for(j = i + 1; j <= k && S[tmp[j]].y - S[tmp[i]].y < d; j++) 
            double d3 = dist(tmp[i], tmp[j]);
            if(d > d3)
                d = d3;
        
    return d;


int main() 
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
        scanf("%lf%lf", &S[i].x, &S[i].y);
    sort(S + 1, S + n + 1, cmpx);
    printf("%.4f\n", merge(1, n));
    return 0;

效率更高的随机算法，但是不能保证正确性因为只取了至多后10个点，假如把这个去掉又有可能被卡成T，不过从旋转角到旋转原点都是完全随机的，多试几次可能就行了，假如不放心的话可以多Calc几次。
假如取至多后20个点的话应该是极限了，假如后20个点都过不了宁愿再重新Calc一次吧。

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

const double PI = acos(-1.0);

struct Point 
    double x, y;
 p[200000 + 5];

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 x0 = -10.0 + 20.0 * RandomDouble(), y0 = -10.0 + 20.0 * RandomDouble(); //随机弄一个旋转原点
    for(int i = 1; i <= n; i++) 
        double x = p[i].x, y = p[i].y, xn, yn;
        xn = (x - x0) * cos(A) - (y - y0) * sin(A) + x0 ;
        yn = (x - x0) * sin(A) + (y - y0) * cos(A) + y0 ;
        p[i].x = xn;
        p[i].y = yn;
    
    sort(p + 1, p + 1 + n, cmp);

    double d = 1e36;
    for(int i = 1; i <= n; i++) 
        for(int j = i + 1; j <= n && j <= i + 10 && p[j].x - p[i].x < d; j++)
            d = min(d, dis(p[i], p[j]));
    
    return d;


int main() 
    srand(time(0));

    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
        scanf("%lf%lf", &p[i].x, &p[i].y);

    printf("%.4f\n", Calc(RandomDouble() * 2.0 * PI));
    return 0;

类似这样？

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

const double PI = acos(-1.0);

struct Point 
    double x, y;
 p[200000 + 5];

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
    for(int i = 1; i <= n; i++) 
        double x = p[i].x, y = p[i].y, xn, yn;
        xn = (x - x0) * cos(A) - (y - y0) * sin(A) + x0 ;
        yn = (x - x0) * sin(A) + (y - y0) * cos(A) + y0 ;
        p[i].x = xn;
        p[i].y = yn;
    
    sort(p + 1, p + 1 + n, cmp);

    for(int i = 1; i <= n; i++) 
        for(int j = i + 1; j <= n && j <= i + 15 && p[j].x - p[i].x < d; j++)
            d = min(d, dis(p[i], p[j]));
    
    return d;


int main() 
    srand(time(0));

    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
        scanf("%lf%lf", &p[i].x, &p[i].y);

    double d = 1e36;
    d = Calc(RandomDouble() * 2.0 * PI, d);
    d = Calc(RandomDouble() * 2.0 * PI, d);
    printf("%.4f\n", d);
    return 0;

卡了一下常数，有很多多余的浮点运算。假如把至多后15个点的限制去掉正确性就可以保证了，但是复杂度可能会爆炸，不过这样子就只需要计算一次就可以了。

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

const double PI = acos(-1.0);

struct Point 
    double x, y;
 p[200000 + 5];

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, xn, yn;
        //p[i].x = (x - x0) * cosA - (y - y0) * sinA + x0;
        //p[i].y = (x - x0) * sinA + (y - y0) * cosA + y0;
        p[i].x = x * cosA - y * sinA + xc;
        p[i].y = x * sinA + y * cosA + yc;
    
    sort(p + 1, p + 1 + n, cmp);

    for(int i = 1; i <= n; i++) 
        for(int j = i + 1; j <= n /*&& j <= i + 15*/ && p[j].x - p[i].x < d; j++)
            d = min(d, dis(p[i], p[j]));
    
    return d;


int main() 
    srand(time(0));

    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
        scanf("%lf%lf", &p[i].x, &p[i].y);

    double d = 1e36;
    d = Calc(RandomDouble() * 2.0 * PI, d);
    //d = Calc(RandomDouble() * 2.0 * PI, d);
    printf("%.4f\n", d);
    return 0;