标签:答案 ace 有一个 tps https 频道 map 相同 位置
两两求距离,将最小生成树的每条边保存到数组,共p-1条边,s个卫星形成s-1条边。
较大s-1条边 使用微型,答案即a[p-s]。
#include <iostream>
#include <memory.h>
#include <string>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <algorithm>
#include <map>
#include <queue>
#include <math.h>
#include <cstdio>
using namespace std;
typedef long long LL;
const int MAXM = 250000+10;
const int MAXN = 500+10;
struct Node
{
double _x,_y;
}node[MAXN];
struct Path
{
int _l,_r;
double _value;
bool operator < (const Path & that)const{
return this->_value < that._value;
}
}path[MAXM];
int Father[MAXN];
double a[MAXN];
int n;
int s,p;
int Get_F(int x)
{
return Father[x] = (Father[x] == x) ? x : Get_F(Father[x]);
}
void Init()
{
for (int i = 1;i <= n;i++)
Father[i] = i;
}
double Get_Len(Node a,Node b)
{
return sqrt((a._x - b._x) * (a._x - b._x) + (a._y - b._y) * (a._y - b._y));
}
int main()
{
cin >> n;
while (n--)
{
memset(a,0,sizeof(a));
cin >> s >> p;
for (int i = 1;i <= p;i++)
Father[i] = i;
for (int i = 1;i <= p;i++)
cin >> node[i]._x >> node[i]._y;
int pos = 0;
for (int i = 1;i <= p;i++)
{
for (int j = i + 1;j <= p;j++)
{
path[++pos]._l = i;
path[pos]._r = j;
path[pos]._value = Get_Len(node[i], node[j]);
}
}
sort(path + 1,path + 1 + pos);
int cnt = 0;
for (int i = 1;i <= pos;i++)
{
int tl = Get_F(path[i]._l);
int tr = Get_F(path[i]._r);
if (tl != tr)
{
Father[tl] = tr;
a[++cnt] = path[i]._value;
}
}
printf("%.2lf\n",a[p-s]);
}
return 0;
}
标签:答案 ace 有一个 tps https 频道 map 相同 位置
原文地址:https://www.cnblogs.com/YDDDD/p/10338933.html