难题再发，请强人指导！！大家多多讨论，谢谢了！！！

问题一：
给定一组随即点的纵横坐标(x[i],y[i]),再给定一个点P，计算P到这组点的最近距离，也就是求一个点Q，
使得P到其他所有点的距离都不小于P到Q的距离，主要是计算的速度，挨个查询效率很低。
问题二：给一组折线断和一个点P，计算P到这组线段的最近距离，也就是求一段线段和这段线段上的一个点Q。
使得P到其他线段上的点的距离都不小于P到Q的距离，并求出这段线段的起点到Q的距离
小弟不知如何优化，查询是太慢了，请指教！！！

typedef struct coordinate
{
double x_cd;
double y_cd;
}* Point;/*Using a struct with two members 'x_cd' and 'y_cd' to descripe a point */
struct Line
{
coordinate StartPoint;
coordinate EndPoint;
};
#define MAXSIZE 10000/* Maximum number of points*/
#define MAXDISTANCE 100000/*maximum distance between points */
#include<iostream>
#include<fstream>
#include<cmath>
using namespace std;
Point Read_data_from(char* filename);/*read coordinate of points from finame and store it in
array PointsArray ,then return the address of the array*/
double Min_dis(Point PointsArray,coordinate KeyPoint);/*Calculate the distance between point 'Keypoint' and the other points,
then return the index of the point who is the nearest point to KeyPoint,meanwhile store the minimum diatance
in the first position of the array as PointsArray[0].y_cd */
int main(int argc,char*argv[])
{
coordinate KeyPoint;
coordinate* Points=Read_data_from("data.txt");/*read data from file and store it in the array */
int size=(Points[0].x_cd--);/*Get the size of Point array */
KeyPoint=Points[size];/*Read the keypoint lastly */
cout<<Min_dis(Points,KeyPoint)<<endl;
cout<<Points[0].y_cd<<endl;
return 0;
}
/***********************************************************************************************************/
Point Read_data_from(char *finame)
{
ifstream read(finame);
if(read==NULL)
{
cout<<"Error,file not exist\n";
exit(1);
}/*If the file can not be opened ,end the programe*/
int counter(0);/*To count the number of points read in*/
Point PointsArray=new coordinate[MAXSIZE];
while(read>>PointsArray[counter+1].x_cd && read>>PointsArray[counter+1].y_cd)
counter++;
PointsArray[0].x_cd=counter;/*We use the first position to store the number of points */
return PointsArray;
}
/*************************************************************************************************************/
double Min_dis(Point PointsArray,coordinate KeyPoint)
{
const length=PointsArray[0].x_cd;/*Get the size of array*/
double x=KeyPoint.x_cd,y=KeyPoint.y_cd;
int index(0);/* the number of the nearest point */
double MinDistance(MAXDISTANCE);/*initailize the minmun distance with a big number */
for(int i=1;i<=length;i++)/*Loop to calculate every distance*/
{
double temX=PointsArray[i].x_cd;
double temY=PointsArray[i].y_cd;
double dis=sqrt((x-temX)*(x-temX)+(y-temY)*(y-temY));
MinDistance= MinDistance > dis ? dis : MinDistance;/*if new distance is small than MinDistance,change
MinDistance with new distance 'dis',and then change the index of the nearest point */
index= MinDistance >= dis ? i:index;
}
PointsArray[0].y_cd=index;
return MinDistance;/*store the index of the nearest point as PointsArray[0].y_cd*/
}
这是我写的，但是从点的平面分布来说，可能有以下的优化方法：
比如：把平面划分成若干区域，应最先在已知点的最近邻近区域内找，找到了当然最近。
还有，就是以已知点为圆心搜索其周围，最先找到的应该为最优解，但是我不好实现！
还请各位指点一二，万分感谢！！！