[洛谷P1991]无线通讯网

题目传送门

题意不难理解，实质就是最小生成树（MST），板子题，这里用的是$Kruskal$。$Prim$做法参考P1265。

#include <bits/stdc++.h>

using namespace std;

#define re register
#define rep(i, a, b) for (re int i = a; i <= b; ++i)
#define repd(i, a, b) for (re int i = a; i >= b; --i)
#define maxx(a, b) a = max(a, b);
#define minn(a, b) a = min(a, b);
#define LL long long
#define inf (1 << 30)

inline int read() {
    int w = 0, f = 1; char c = getchar();
    while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar();
    while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar();
    return w * f;
}

const int maxn = 500 + 5;

#define sqr(x) ((x) * (x))

struct Edge {
    int u, v;
    double w;
} e[maxn * maxn];
bool cmp(Edge a, Edge b) { return a.w < b.w; }

int f[maxn];
int fset(int x) { return f[x] == x ? x : f[x] = fset(f[x]); }
void merge(int x, int y) { f[fset(x)] = fset(y); }

int n, x[maxn], y[maxn], m;

int main() {
    m = read(), n = read();

    rep(i, 1, n) x[i] = read(), y[i] = read();

    rep(i, 1, n)
        rep(j, 1, n)
            e[i*n+j-n] = (Edge){i, j, sqrt(sqr(x[i]-x[j]) + sqr(y[i]-y[j]))};

    sort(e+1, e+n*n+1, cmp);

    int p = 1;
    rep(i, 1, n) f[i] = i;

    while (n-- > m) {
        while (fset(e[p].u) == fset(e[p].v)) p++;
        merge(e[p].u, e[p].v);
    }

    printf("%.2lf", e[p].w);

    return 0;
}