杭电1007题

普通方法，超时了。

#include<iostream>#include <iomanip>#include<math.h>#include<vector>using namespace std;class Point{public:double douX;double douY;Point():douX(0),douY(0){}Point(double x, double y):douX(x),douY(y){}};int main(){int iPointNum;double dX, dY;while(cin >> iPointNum){if(0 == iPointNum){return 0;}else{vector<Point> vecPoint(iPointNum); double douMinDistance = 0;double douDistance = 0;for(int i = 0; i < iPointNum; i++){cin >> dX >> dY;Point p(dX, dY);vecPoint[i] = p;}//for_loopfor(int j = 0; j < iPointNum; j++){for(int k = j + 1; k < iPointNum; k++){double douXDistence = fabs(vecPoint[k].douX - vecPoint[j].douX);double douYDistence = fabs(vecPoint[k].douY - vecPoint[j].douY);douDistance = sqrt(pow(douXDistence, 2) + pow(douYDistence, 2));if(douDistance < douMinDistance){douMinDistance = douDistance;}}//for_loop}//for_loopcout << fixed << setprecision(2) << douMinDistance << endl;}//if_loop}//while_loopreturn 0;}

模仿网上一个分治法，还是超时了。

#include<iostream>#include <iomanip>#include<math.h>#include<vector>#include<algorithm>using namespace std;class Point{public:double douX;double douY;Point():douX(0),douY(0){}Point(double x, double y):douX(x),douY(y){}};double P2PDistance(Point pa, Point pb){double douXDistence = fabs(pa.douX - pb.douX);double douYDistence = fabs(pa.douY - pb.douY);double douDistance = sqrt(pow(douXDistence, 2) + pow(douYDistence, 2));return douDistance;}double MinDistance(double a, double b){return a < b ? a : b;}bool CompByX(Point pa, Point pb){if(pa.douX < pb.douX){return true;}else{return false;}}bool CompByY(Point pa, Point pb){if(pa.douY < pb.douY){return true;}else{return false;}}double GetMinDistance(int begin, int end, vector<Point>& vecPoint){if(begin == end){return 0;}else if(begin + 1 == end){return P2PDistance(vecPoint[begin], vecPoint[end]);}else if(begin + 2 == end){double dist1 = P2PDistance(vecPoint[begin], vecPoint[begin+1]);  double dist2 = P2PDistance(vecPoint[begin+1], vecPoint[begin+2]);  double dist3 = P2PDistance(vecPoint[begin], vecPoint[begin+2]);  return MinDistance(MinDistance(dist1,dist2),dist3);}int mid = (begin + end)/2;int iMinDist = MinDistance( GetMinDistance(begin, mid, vecPoint), GetMinDistance(mid + 1, end, vecPoint) );       //vector里面的元素已经X排序vector<Point> vecPointX;for(int i = begin; i < end; i++){if(fabs(vecPoint[i].douX - vecPoint[mid].douX) < iMinDist){vecPointX.push_back(vecPoint[i]);                                                                //X距离小于iMinDist，有可能}}//for_loop                                                                 vector<Point> vecPointY;sort(vecPointX.begin(), vecPointX.end(), CompByY);mid = (vecPointX.size()) / 2;for(int i = 0; i < vecPointX.size(); i++){if(fabs(vecPointX[i].douY - vecPointX[mid].douY) < iMinDist){vecPointY.push_back(vecPointX[i]);                                                                //Y距离小于iMinDist，有可能}}//for_loopfor(int i = 0; i < vecPointY.size(); i++)                                                                           //查看上述那些“可能”是否是会出现{double dist = P2PDistance(vecPointY[i], vecPointX[mid]);if(dist < iMinDist){iMinDist = dist;}}//for_loopreturn iMinDist;}int main(){int iPointNum;double dX, dY;while(cin >> iPointNum){if(0 == iPointNum){return 0;}else{vector<Point> vecPoint(iPointNum); double douMinDistance = 0;for(int i = 0; i < iPointNum; i++){cin >> dX >> dY;Point p(dX, dY);vecPoint[i] = p;}//for_loop sort(vecPoint.begin(), vecPoint.end(), CompByX);                                //对X进行排序，进入GetMinDistance使用 douMinDistance = GetMinDistance(0, vecPoint.size() - 1, vecPoint); cout << fixed << setprecision(2) << douMinDistance << endl;}//if_loop}//while_loopreturn 0;}

题目：点击打开链接    ----   最小套圈问题

参考答案： 点击打开链接        ----  分治法求解

传统观的分治法，是将一个大问题分解成同样的小问题求解。这个里面有点状况，不能完全套用传统的方法。将一堆点按X坐标排序后，找出某个基准点将这堆点分为左右两个集合，大问题（寻找这堆点的最小距离）就分解为找左右两个集合里面最小问题，还加一个特殊情况（可能是左边某点和右边某点的距离是最小的）。这个题就是看你怎么处理这个特殊情况了。