「最小生成树」[HAOI2006]聪明的猴子

[HAOI2006]聪明的猴子

原题链接 [HAOI2006]聪明的猴子

题目大意

给你(n)个数，代表最多可以走多长的边权，再给你(m)个坐标，每个坐标给(x, y)两个值，求在(n)中有多少值能够走完(m)个这些坐标

题目题解

很水的一道题，只是熟悉下模板 就是坐标两两连线，然后用最小生成树乱搞一下

(最小生成树模板出错改了一个小时 quq)

//#define fre yes

#include <cmath>
#include <cstdio>
#include <iostream>
#include <algorithm>
#define int long long

const int N = 10005;
int par[N];
int b[N], dx[N], dy[N];

struct Node {
    int x, y, k;
} edgee[N * 60];

bool cmp(Node x, Node y) {
    return x.k < y.k;
}

int find(int x) {
    if(x == par[x]) return par[x];
    else return par[x] = find(par[x]);
}

int tnt = 0;
signed main() {
    static int n, m;
    scanf("%lld", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%lld", &b[i]);
    }

    scanf("%lld", &m);
    for (int i = 1; i <= m; i++) {
        int x, y;
        scanf("%lld %lld", &x, &y);
        dx[i] = x; dy[i] = y;
        par[i] = i;
    }

    int cnt = 0;
    for (int i = 1; i < m; i++) {
        for (int j = i + 1; j <= m; j++) {
            edgee[++cnt].x = i;
            edgee[cnt].y = j;
            edgee[cnt].k = sqrt(((dx[i] - dx[j]) * (dx[i] - dx[j])) + ((dy[i] - dy[j]) * (dy[i] - dy[j])));
        }
    }

    std::sort(edgee + 1, edgee + 1 + cnt, cmp);

    int num = 0;
    for (int i = 1; i <= cnt; i++) {
        int x = find(edgee[i].x);
        int y = find(edgee[i].y);
        if(x == y) continue;

        par[x] = y;
        tnt = std::max(tnt, edgee[i].k);
        num++;
        if(num == m - 1) break;
    }

    int ans = 0;
    for (int i = 1; i <= n; i++) {
        if(b[i] >= tnt) ans++;
    } printf("%lld
", ans);
    return 0;
}