P4047 [JSOI2010]部落划分题解

P4047 [JSOI2010]部落划分

题意：给定一些点，要把这些点分为k个部落，问最近两个部落间的最远距离是多少。

解法：二分答案+并查集检测鸭！二分的时候要注意精度问题噢，eps开到1e-3会有一个点过不去。开始看错题意想半天。

代码：

#include  
#include  
#include  
#include  
#include  
#include  
#include  
#include  
#include  
#include
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define _for(n,m,i) for (register int i = (n); i < (m); ++i)
#define _rep(n,m,i) for (register int i = (n); i <= (m); ++i)
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
#define PI acos(-1)
#define eps 1e-4
#define rint register int
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)
using namespace std;
typedef long long LL;
typedef pair<LL, int> pli;
typedef pair<int, int> pii;
typedef pair<double, int> pdi;
typedef pair<LL, LL> pll;
typedef pair<double, double> pdd;
typedef map<int, int> mii;
typedef map<LL, int> mli;
typedef map<char, int> mci;
typedef map<string, int> msi;
template<class T>
void read(T &res) {
     
  int f = 1; res = 0;
  char c = getchar();
  while(c < '0' || c > '9') {
      if(c == '-') f = -1; c = getchar(); }
  while(c >= '0' && c <= '9') {
      res = res * 10 + c - '0'; c = getchar(); }
  res *= f;
}
const int ne[8][2] = {
     1, 0, -1, 0, 0, 1, 0, -1, -1, -1, -1, 1, 1, -1, 1, 1};
const LL INF = 1e18;
const int N = 1010;
const LL Mod = 1e9+7;
const int M = 110;
int n, k;
pdd p[N];
double tu[N][N];
double dis(pdd p1, pdd p2) {
     
  return sqrt((p1.first-p2.first)*(p1.first-p2.first) + (p1.second-p2.second)*(p1.second-p2.second));
}
int f[N];
int Find(int x) {
     
  return f[x] == x ? x : f[x] = Find(f[x]);
}
int check(double x) {
     
  _rep(1, n, i) f[i] = i;
  int fi, fj;
  _rep(1, n, i) _for(1, i, j) {
     
    if(tu[i][j] >= x) continue;
    fi = Find(i); fj = Find(j);
    if(fi != fj) f[fi] = fj;
  }
  int cnt = 0;
  _rep(1, n, i) {
     
    if(Find(i) == i) ++cnt;
  }
  //cout << x << " " << cnt << endl;
  return cnt >= k;
}
int main() {
     
  scanf("%d%d", &n, &k);
  double mx = 0;
  _rep(1, n, i) {
     
    scanf("%lf%lf", &p[i].first, &p[i].second);
    _for(1, i, j) tu[i][j] = tu[j][i] = dis(p[i], p[j]), mx = max(mx, tu[i][j]);
  }
  double l = 0, r = mx, mid, ans = mx;
  while(r-l>eps) {
     
    mid = (l + r) / 2.0;
    if(check(mid)) ans = mid, l = mid;
    else r = mid;
  }
  printf("%.2lf\n", ans);
  return 0;
}