【BZOJ2429】聪明的猴子

题目链接：https://www.lydsy.com/JudgeOnline/problem.php?id=2429

---

比较简单的一道题目。。。

从一棵树出发蹦到其他所有树，实际上，把n-1条边就可以将这n个点连通，而最小生成树的最大权值又是所有生成树中最小的，因此我们需要构造一棵最小生成树，然后将m个猴子的跳跃距离和MST的最大边权比较，统计答案即可。

#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

const int maxm = 505, maxn = 1e3 + 5;

int monkey[maxm], treex[maxn], treey[maxn];

struct Edge {
    int u, v;
    double w;
    bool operator < (const Edge& rhs) const {
        return w < rhs.w;
    }
} edge[maxn * maxn / 2];

inline int pw2(int x) {
    return x * x;
}

inline double dis(int i, int j) {
    return sqrt(pw2(treex[i] - treex[j]) + pw2(treey[i] - treey[j]));
}

int m, n, eid, fa[maxn];

int dj_find(int i) {
    if (i == fa[i]) return i;
    else return fa[i] = dj_find(fa[i]);
}

inline void dj_merge(int a, int b) {
    fa[dj_find(a)] = dj_find(b);
}

inline double kruskal() {
    for (int i = 1; i <= n; ++i) fa[i] = i;
    sort(edge + 1, edge + eid + 1);
    int cnt = 0;
    for (int i = 1; i <= eid; ++i) {
        int u = edge[i].u, v = edge[i].v, w = edge[i].w;
        if (dj_find(u) != dj_find(v)) {
            dj_merge(u, v);
            if (++cnt == n - 1) return w;
        }
    }
    return edge[eid].w;
}

int main() {
    scanf("%d", &m);
    for (int i = 1; i <= m; ++i) scanf("%d", &monkey[i]);
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) scanf("%d%d", &treex[i], &treey[i]);
    for (int i = 1; i < n; ++i)
        for (int j = i + 1; j <= n; ++j) {
            edge[++eid].u = i;
            edge[eid].v = j;
            edge[eid].w = dis(i, j);
        }
    int ans = 0;
    double me = kruskal();
    for (int i = 1; i <= m; ++i)
        if (monkey[i] >= me) ++ans;
    printf("%d", ans);
    return 0;
}