[AcWing 4266] 无线网络

特定条件合并集合

并查集

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点击查看代码

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 1e6 + 10;

int n, d;
int p[N];
bool open[N];

struct Node
{
    int x, y;
} s[N];

int find(int x)
{
    if (p[x] != x)
        p[x] = find(p[x]);
    return p[x];
}

void merge(int a, int b)
{
    int pa = find(a), pb = find(b);
    p[pa] = pb;
}

bool check(int a, int b)
{
    int dx = s[a].x - s[b].x;
    int dy = s[a].y - s[b].y;
    return dx * dx + dy * dy <= d * d;
}

void solve()
{
    cin >> n >> d;
    for (int i = 1; i <= n; i ++)
        cin >> s[i].x >> s[i].y;
    for (int i = 1; i <= n; i ++)
        p[i] = i;
    char op;
    while (cin >> op) {
        int p, q;
        if (op == 'O') {
            cin >> p;
            open[p] = true;
            for (int i = 1; i <= n; i ++) {
                if (open[i] && check(p, i))
                    merge(i, p);
            }
        }
        else {
            cin >> p >> q;
            if (find(p) == find(q))
                cout << "SUCCESS" << endl;
            else
                cout << "FAIL" << endl;
        }
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    solve();

    return 0;
}

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1. 当一个电脑开机时，枚举其他电脑，将能够合并的进行合并，可以进行合并的条件：
   ① 两个电脑都开机
   ② 两个电脑之间的欧氏距离小于等于 $d$