问题:
给定平面上N个点的坐标,找出距离最近的两个点。
解法:
我们先对N个点的x坐标进行排序,排序我们使用最坏复杂度O(n*logn)的快速排序方法,在排序的过程中minDifferent会递归计算出左右两边的最小距离,再用其中的较小值minum得到以中位数点附近的带状区域[p[median+1].x-median, p[median].x+median],对带状区域的点按照y坐标排序,对带状区域的每个点只需计算最多7个点,就能得到所有可能小于minum的点对。
#include <iostream> #include <vector> #include <algorithm> #include <cmath> using namespace std; // 顶点信息 struct Point { double m_x, m_y; Point():m_x(0.0),m_y(0.0) {} Point(double x, double y):m_x(x),m_y(y){} bool operator==(const Point& p) const {return m_x==p.m_x && m_y==p.m_y;} }; ostream& operator<<(ostream& os, const Point& p) { return os << "(" << p.m_x << "," << p.m_y << ")"; } // 插入排序 template<class T, class Pr> void insert_sort(vector<T> &vec, int l, int r, Pr pred) { int i, j; for (i=l+1; i<=r; i++) { T tmp = vec[i]; for (j=i-1; j>=l && pred(tmp,vec[j]); j--) vec[j+1]=vec[j]; vec[j+1] = tmp; } } // 找到key所在的位置 template<class T> int get_position(vector<T> &vec, int l, int r, T key) { for (int i=l; i<=r; i++) if (key == vec[i]) return i; return -1; } // 按第一个元素对vec进行划分 template<class T, class Pr> int partition(vector<T> &vec, int l, int r, Pr pred) { int i, j; for (i=l+1,j=l; i<=r; i++) { if (pred(vec[i],vec[l])) { ++j; swap(vec[i],vec[j]); } } swap(vec[j],vec[l]); return j; } // 顺序统计得到第k个元素的值 template<class T, class Pr> T select(vector<T> &vec, int l, int r, int k, Pr pred) { int n = r-l+1; if (n==1) { if (k!=0) printf("Out of Boundary!\n"); return vec[l]; } // 找中位数的中位数作为分割点 int cnt = n/5; int tcnt = (n+4)/5; int rem = n%5; vector<T> group(tcnt); int i, j; for (i=0,j=l; i<cnt; i++,j+=5) { insert_sort(vec, j, j+4, pred); group[i] = vec[j+2]; } if (rem) { insert_sort(vec, j, j+rem-1, pred); group[i] = vec[j+(rem-1)/2]; } T key = select(group, 0, tcnt-1, (tcnt-1)/2, pred); // 找到分割点的位置 int key_pos = get_position(vec, l, r, key); swap(vec[key_pos], vec[l]); // 用分割点对数组进行划分,小的在左边,大的在右边 int pos = partition(vec, l, r, pred); int x = pos - l; if (x == k) return key; else if (x < k) return select(vec, pos+1, r, k-x-1, pred); else return select(vec, l, pos-1, k, pred); } // 计算点a和b的距离 double dist(const Point& a, const Point& b) { double x = a.m_x-b.m_x; double y = a.m_y-b.m_y; return sqrt(x*x+y*y); } bool cmpX(const Point& a, const Point& b) { return a.m_x < b.m_x; } bool cmpY(const Point& a, const Point& b) { return a.m_y < b.m_y; } double minDifferent(vector<Point> p, int l, int r, vector<Point> &result) { // 按中位数进行划分后的子区域的元素个数都会减小到2或3,不会再到1 if ((r-l+1)==2) { result[0] = p[l]; result[1] = p[r]; if (cmpX(p[r],p[l])) swap(p[l], p[r]); return dist(p[l], p[r]); } if ((r-l+1)==3) { insert_sort(p, l, r, cmpX); double tmp1 = dist(p[l], p[l+1]); double tmp2 = dist(p[l+1], p[l+2]); double ret = min(tmp1, tmp2); if (tmp1 == ret) { result[0] = p[l]; result[1] = p[l+1]; } else { result[0] = p[l+1]; result[1] = p[l+2]; } return ret; } // 大于3个点的情况 int mid = (r+l)>>1; Point median = select(p, l, r, mid-l, cmpX); vector<Point> res1(2), res2(2); double min_l = minDifferent(p, l, mid, res1); double min_r = minDifferent(p, mid+1, r, res2); double minum = min(min_l, min_r); if (minum == min_l) { result[0] = res1[0]; result[1] = res1[1]; } else { result[0] = res2[0]; result[1] = res2[1]; } // 对[p[mid+1]-minum, p[mid]+minum]的带状区域按y排序 vector<Point> yvec; int i, j; for (i=mid+1; i<=r; i++) if (p[i].m_x - p[mid].m_x < minum) yvec.push_back(Point(p[i])); for (i=mid; i>=l; i--) if (p[mid+1].m_x - p[i].m_x < minum) yvec.push_back(Point(p[i])); sort(yvec.begin(), yvec.end(), cmpY); for (i=0; i<yvec.size(); i++) { // 至多只有与其后最多7个点的距离会小于minum for (j=i+1; j<yvec.size() && yvec[j].m_y-yvec[i].m_y<minum && j<=i+7; j++) { double delta = dist(yvec[i],yvec[j]); if (delta < minum) { minum = delta; result[0] = yvec[i]; result[1] = yvec[j]; } } } return minum; } int main() { int n, i, j, x, y; vector<Point> result(2); vector<Point> input; cin >> n; for (i=0; i<n; i++) { cin >> x; cin >> y; input.push_back(Point(x,y)); } double minum = minDifferent(input, 0, input.size()-1, result); cout << "nearest point: " << result[0] << " and " << result[1] << endl; cout << "distance: " << minum << endl; return 0; }POJ 3714 问题:
平面上有两类点,计算属于不同类的顶点对的最小值。
解法:参考http://blog.csdn.net/smsmn/article/details/5963487
算法思想与上面基本相同,但编程方式上进行了改变,更好理解。下面的代码写法上完全按照参考代码的思路,只是将数组操作改为vector,但提交到POJ 3714上却会TLE,看来动态分配空间所占用的时间也不小。所以要想AC,请使用参考代码。
#include <iostream> #include <vector> #include <algorithm> #include <cmath> using namespace std; const double INF = 1e100; struct Point { double x, y; int flag; // 顶点的类别 Point(){} Point(double xx, double yy):x(xx),y(yy){} }; vector<Point> p; bool cmp1(const Point& a, const Point& b) { return a.x < b.x; } bool cmp2(int a, int b) { return p[a].y < p[b].y; } double dist(const Point& a, const Point& b) { double xx = a.x - b.x; double yy = a.y - b.y; return sqrt(xx*xx+yy*yy); } // 输出属于不同类的顶点对的最小值 double min_dist(vector<Point> p, int left, int right) { int mid = (left+right)>>1, i,j; if (left>=right) return INF; for (i=mid; i>=left && p[mid].x<=p[i].x; i--); double minum = min_dist(p, left, i); for (i=mid; i<=right && p[i].x<=p[mid].x; i++); minum = min(minum, min_dist(p, i, right)); vector<int> yp; for (i=mid; i>=left && p[mid].x-p[i].x<minum; i--) yp.push_back(i); for (i=mid+1; i<=right && p[i].x-p[mid].x<minum; i++) yp.push_back(i); // 这个方法非常巧妙,直接对顶点索引进行排序,减少了空间使用, // 代码上也更加简洁 sort(yp.begin(), yp.end(), cmp2); for (i=0; i<yp.size(); i++) for (j=i+1; j<yp.size() && p[yp[j]].y-p[yp[i]].y<minum; j++) // 主要的不同之处,产生最小距离的点对必须属于不同类别 if (p[yp[j]].flag != p[yp[i]].flag) minum = min(minum, dist(p[yp[j]], p[yp[i]])); return minum; } int main() { int i,j,test; cin >> test; while (test--) { int xx,yy,n; cin >> n; for (i=0; i<n; i++) { cin >> xx >> yy; p.push_back(Point(xx,yy)); p[i].flag = 1; } for (; i<2*n; i++) { cin >> xx >> yy; p.push_back(Point(xx,yy)); p[i].flag = 2; } // 按照x坐标对点集进行排序 sort(p.begin(), p.end(), cmp1); printf("%.3lf\n", min_dist(p, 0, p.size()-1)); } }