POJ 2560 Freckles
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 5217 | Accepted: 2706 |
Description
Consider Dick's back to be a plane with freckles at various (x,y) locations. Your job is to tell Richie how to connect the dots so as to minimize the amount of ink used. Richie connects the dots by drawing straight lines between pairs, possibly lifting the pen between lines. When Richie is done there must be a sequence of connected lines from any freckle to any other freckle.
Input
Output
Sample Input
3 1.0 1.0 2.0 2.0 2.0 4.0
Sample Output
3.41
Source
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <cmath>
#define Y 103
using namespace std;
struct node
{
double x,y;
};
node c[Y];
double p[Y][Y],Max;
int N;
bool b[Y];
double dis(double &x1,double &y1,double &x2,double &y2)
{
return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
void prim()
{
int i,j,t;
for(i=1;i<=N;i++)
b[i]=0;
b[1]=1;
t=N-1;
double min,s=0;
while(t--)
{ min=Max;
for(i=2;i<=N;i++)
if(!b[i]&&p[1][i]<min)
{
j=i;
min=p[1][i];
}
s+=min;
b[j]=1;
for(i=2;i<=N;i++)
if(!b[i]&&p[1][i]>p[j][i])
p[1][i]=p[j][i];
}
printf("%.2lf\n",s);
}
int main()
{
while(scanf("%d",&N)!=EOF)
{
for(int i=1;i<=N;i++)
scanf("%lf%lf",&c[i].x,&c[i].y);
Max=0;
for(int j=1;j<N;j++)
for(int k=j+1;k<=N;k++)
{
p[j][k]=dis(c[j].x,c[j].y,c[k].x,c[k].y);
p[k][j]=p[j][k];
Max=Max>p[j][k]?Max:p[j][k];
}
prim();
}
return 0;
}