HDU-1162-Eddy's picture

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;

const int maxn = 100+10;
int far[maxn];
int n, m;
int k;
struct Edge
{
const bool operator<(Edge &n)const
{
return value<n.value;
}
int start;
int end;
double value;
}edge[maxn*maxn];

bool cmp(struct Edge &a, struct Edge &b)
{
return a.value<b.value;
}

void init()
{
int i, j;
for(i=0; i<=n; i++)
far[i] = i;
}

int find(int a)
{
if(far[a]!=a)
far[a] = find(far[a]);
return far[a];
}

void Union(int a, int b)
{
a = find(a);
b = find(b);
if(a != b)
far[a] = b;
}

double kruskal()
{
init();
sort(edge,edge+n*(n-1)+1,cmp);
double sum = 0;
int count = 0;
for(int i=1; i<=n*(n-1) && count<n; i++)
{
int a=edge[i].start;
int b=edge[i].end;
if(find(a)!=find(b))
{
Union(a, b);
sum+=edge[i].value;
count++;
}
}
return sum;
}

int main()
{
double a[maxn], b[maxn];
while(cin>>n)
{
//m = n*(n-1)/2;

for(int i=1; i<=n; i++)
{
scanf("%lf%lf",&a[i], &b[i]);  //因为是无向图，所以要两次赋值
}
k=1;
double c;
for(int i=1; i<n; i++)
{
for(int j=i+1; j<=n; j++)
{
//if(i != j)
//{
c = sqrt((a[i]-a[j])*(a[i]-a[j])+(b[i]-b[j])*(b[i]-b[j]));
edge[k].start = i;
edge[k].end = j;
edge[k++].value = c;
edge[k].start = j;
edge[k].end = i;
edge[k++].value = c;
//}
}
}
printf("%.2lf\n", kruskal());
}
return 0;
}