寻找最近点对

fcpp.h

#ifndef FCCP_H
#define FCCP_H

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <malloc.h>

typedef int ElemType;

ElemType position(ElemType a[],ElemType b[],int start,int end);

void swap(ElemType &a,ElemType &b)
{
    ElemType temp;
    temp = a;
    a = b;
    b = temp;
}

void sortTwo(ElemType a[],ElemType b[],int start,int end)//使用快速排序实现对a的排序，其中保持b中元素位置相对于a不变
{
    if(start < end)
    {
        int pos = position(a,b,start,end);
        sortTwo(a,b,start,pos-1);
        sortTwo(a,b,pos+1,end);
    }
}

ElemType position(ElemType a[],ElemType b[],int start,int end)//快速排序求划分位置方法
{
    ElemType temp = a[start],temp2 = b[start];
    int i = start,j = end;

    while(i < j)
    {
        while(j > i && a[j] >= temp) 
        {
            j--;
        }
        if(i < j)
        {
            a[i] = a[j];
            b[i] = b[j];
        }
        while(j > i && a[i] <= temp)
        {
            i++;
        }
        if(i < j)
        {
            a[j] = a[i];
            b[j] = b[i];
        }
    }
    a[i] = temp;
    b[i] = temp2;
    return i;
}

void sort(ElemType a[],ElemType b[],int n)//对按照a排序后中a值相同的情况再按照b中元素的大小进行排序
{
    for(int i = 0;i < n-1;)
    {
        int s,e;
        while(a[i] != a[i+1] && i < n-1)
        {
            i++;
        }
        s = i;
        while(a[i] == a[i+1] && i < n-1)
        {
            i++;
        }
        e = i;
        for(int j = s;j <= e;j++)
        {
            for(int m = e-s+j;m > j;m--)
            {
                if(b[m] < b[m-1])
                {
                    swap(b[m],b[m-1]);
                }
            }
        }
        i++;
    }
}

//---fccp是求解n个点中距离最小的2个点的距离和点的坐标
void fccp(int start,int end,ElemType x[],ElemType y[],ElemType idn[],double &minLen,ElemType &px,ElemType &py,ElemType &qx,ElemType &qy)
{
    int m = end-start+1;//每一步都要判断准确，是大的减去小的才行
    qx = qy = px = py = 0;
    if(m <= 3)//m<=3的情况
    {
        minLen = 10000;
        for(int i = 0;i < m-1;i++)
        {
            for(int j = i+1;j <= m-1;j++)
            {
                int ylen = y[j]-y[i];
                int xlen = x[idn[j]] - x[idn[i]];
                xlen = xlen > 0 ? xlen : -xlen;
                double mlen = sqrt((double)(ylen*ylen+xlen*xlen));
                if(minLen > mlen)
                {
                    minLen = mlen;
                    px = x[idn[i]];
                    py = y[i];
                    qx = x[idn[j]];
                    qy = y[j];
                }
            }
        }
    }
    else//m>=4的情况
    {
        ElemType *YL,*IDNL,*YR,*IDNR;

        int r = m/2;//右边数组长度
        int l = m-r;//左边数组长度
        int lengthL = 0;
        int lengthR = 0;

        YL = (ElemType *)malloc(sizeof(ElemType)*l);
        IDNL = (ElemType *)malloc(sizeof(ElemType)*l);
        YR = (ElemType *)malloc(sizeof(ElemType)*r);
        IDNR = (ElemType *)malloc(sizeof(ElemType)*r);

        for(int i = 0;i < m;i++)
        {
            if(idn[i] <= start+l-1)
            {
                YL[lengthL] = y[i];
                IDNL[lengthL] = idn[i];
                lengthL++;
            }
            else
            {
                YR[lengthR] = y[i];
                IDNR[lengthR] = idn[i];
                lengthR++;
            }
        }

        double mlenl,mlenr;
        ElemType qxl,qxr,qyl,qyr,pxl,pxr,pyl,pyr;

        fccp(start,start+l-1,x,YL,IDNL,mlenl,pxl,pyl,qxl,qyl);
        fccp(start+l,end,x,YR,IDNR,mlenr,pxr,pyr,qxr,qyr);

        if(mlenl < mlenr)
        {
            minLen = mlenl;
            px = pxl;
            py = pyl;
            qx = qxl;
            qy = qyl;
        }
        else
        {
            minLen = mlenr;
            px = pxr;
            py = pyr;
            qx = qxr;
            qy = qyr;
        }

        double xdiv = (double)(x[start+l-1]+x[start+l])/2;//求分割线

        ElemType *YC,*IDNC;
        YC = (ElemType *)malloc(sizeof(ElemType)*m);
        IDNC = (ElemType *)malloc(sizeof(ElemType)*m);
        int lengthC = 0;

        for(int i = 0;i < m;i++)
        {
            if(x[idn[i]] > xdiv-minLen && x[idn[i]] < xdiv+minLen)
            {
                YC[lengthC] = y[i];
                IDNC[lengthC] = idn[i];
                lengthC++;
            }
        }

        if(lengthC <= 0)
        {
            return;
        }

        for(int i = 0;i < lengthC-1;i++)
        {
            ElemType ypres = y[i];
            ElemType xpres = x[IDNC[i]];
            int u = (i+4) > lengthC-1 ? lengthC-1 : (i+4);//u是检查的最大下标
            int v = i+1;
            while(v <= u && YC[v]-ypres < minLen)
            {
                ElemType tempYlen = YC[v]-ypres;
                ElemType tempXlen = x[IDNC[v]] - xpres;
                tempXlen = tempXlen>0 ? tempXlen : -tempXlen;
                double tempLen = sqrt((double)(tempXlen*tempXlen+tempYlen*tempYlen));
                if(tempLen < minLen)
                {
                    minLen = tempLen;
                    px = xpres;
                    py = ypres;
                    qx = x[IDNC[v]];
                    qy = YC[v];
                }
                v++;
            }
        }
    }
}

#endif

main.cpp

#include "fcpp.h"

int main()
{
    //这里先测试数据可以重复，x相同的情况下按照y从小到大排列
    int x[8] = {1,9,3,4,6,7,2,8};
    int y[8] = {5,6,8,12,1,7,4,9};
    int idn[8] = {0,1,2,3,4,5,6,7};

    sortTwo(x,y,0,7);
    sort(x,y,8);

    /*
    for(int i = 0;i < 8;i++)
    {
        printf("%d ",x[i]);
    }
    printf("\n");
    for(int i = 0;i < 8;i++)
    {
        printf("%d ",y[i]);
    }
    printf("\n");
    for(int i = 0;i < 8;i++)
    {
        printf("%d ",idn[i]);
    }
    printf("\n");
    */

    sortTwo(y,idn,0,7);

    /*
    for(int i = 0;i < 8;i++)
    {
        printf("%d ",x[i]);
    }
    printf("\n");
    for(int i = 0;i < 8;i++)
    {
        printf("%d ",y[i]);
    }
    printf("\n");
    for(int i = 0;i < 8;i++)
    {
        printf("%d ",x[idn[i]]);
    }
    printf("\n");
    */

    ElemType px,py,qx,qy;
    double minLen;

    fccp(0,7,x,y,idn,minLen,px,py,qx,qy);

    printf("the shortest length between the two points is : %lf\n\n",minLen);
    printf("the point is ( %d , %d ) and ( %d , %d )",px,py,qx,qy);

    system("pause");
    return 0;
}