http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=98&page=show_problem&problem=88
題意:
給出N個三維數,對於每個點,找出與其它點最小的距離,如距離小於1,則1,0,0,0,0,0,0,0,0,0,0←0,0,0,0,0,0,0,0,0,0。
每個點都得檢索出來,並統計起來。
解題:
如題意,直接保存後開始計算距離。每個點都只找最小距離(與其它點的距離),再統計。(如,對於A點,有到點B距離2.5,有到點C距離3.5,則2.5有效)
寫了個類,複習一下重載~雖然重載了,可發覺用不上。。。當複習好了。
#include <iostream>
#include <iomanip>
#include <vector>
#include <cmath>
#include <string.h>
#include <stdio.h>
using namespace std;
// #define LOCAL_TEST
class CPoint
{
public:
int x;
int y;
int z;
CPoint (int x = 0, int y = 0, int z = 0)
{
this->x = x;
this->y = y;
this->z = z;
}
double fDistance(CPoint p)
{
return sqrt( (double) ((this->x-p.x)*(this->x-p.x)
+ (this->y-p.y)*(this->y-p.y) + (this->z-p.z)*(this->z-p.z) ) );
}
double operator - (CPoint p)
{
return fDistance(p);
}
friend istream &operator << (istream& input, CPoint& p);
};
istream &operator << (istream& input, CPoint& p)
{
input >>p.x >>p.y >>p.z;
return input;
}
int main()
{
#ifdef LOCAL_TEST
freopen("f:\\in.txt", "r", stdin);
freopen("f:\\out.txt", "w+", stdout);
#endif
CPoint p;
int a, b, c;
vector <CPoint> vPt;
vector <double> vDis;
int szCount[12];
memset(szCount, 0, sizeof(szCount));
// get all input data
while ( cin >>a >> b >>c )
{
if ( a == 0 && b == 0 && c == 0 )
break;
CPoint pTmp(a, b, c);
vPt.push_back(pTmp);
} // end while
// calc all distance while index'i!=k
for ( int i=0; i<vPt.size(); i++ )
{
double dis = 11;
// get nearest distance
for ( int k=0; k<vPt.size(); k++ )
if ( i != k && vPt[i] - vPt[k] < dis )
dis = vPt[i] - vPt[k];
if ( dis < 10 + 1e-7 )
{ // count corresponding
int index = (int)(dis) + 1;
szCount[index]++;
} // end if
} // end for
// format output
for ( int i=1; i<=10; i++ )
cout <<setiosflags(ios::right) <<setw(4) <<szCount[i];
cout <<'\n';
return 0;
}