OpenJ_Bailian - 1751

OpenJ_Bailian - 1751

题解：最小生成树问题，Kruskal算法

已经帮你建好的边就不用再建了，直接合并，当然我们这一题需要将给的坐标转化成边的，然后我们如何输出建哪几条路呢？我们知道我们合并的时候其实就是建路的过程，所有我们可以修改一下merge函数，增加一维参数op，代表是否输出从哪到哪建路

#include <bits/stdc++.h>
#define Zeoy std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0)
#define all(x) (x).begin(), (x).end()
#define endl '\n'
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-9;
const int N = 1000;
int n, m, tot;                      //tot表示边数
int fa[N];
void init()
{
    for (int i = 1; i <= n; ++i)
        fa[i] = i;
}
int find(int x)
{
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}
void merge(int x, int y, int op)      //op代表输出从哪到哪建路
{
    int fx = find(x);
    int fy = find(y);
    if (fx != fy)
    {
        fa[fx] = fy;
        if (op)
            cout << x << " " << y << endl;
    }
}
struct node
{
    int u, v;
    double w;
    bool operator<(const node &t) const
    {
        return w < t.w;
    }
} edge[N * N];
pii a[N];
int main(void)
{
    Zeoy;
    int t = 1;
    // cin >> t;
    while (t--)
    {
        cin >> n;
        init();
        for (int i = 1; i <= n; ++i)
        {
            cin >> a[i].first >> a[i].second;
        }
        cin >> m;
        for (int i = 1, u, v; i <= m; ++i)
        {
            cin >> u >> v;
            merge(u, v, 0);             //已经建好的路直接合并就好了
        }
        tot = 1;
        for (int i = 1; i <= n; ++i)
        {
            for (int j = 1; j <= n; ++j)
            {
                if (j != i)
                {
                    double w = sqrt((a[i].first - a[j].first) * (a[i].first - a[j].first) + (a[i].second - a[j].second) * (a[i].second - a[j].second));
                    edge[tot].u = i;
                    edge[tot].v = j;
                    edge[tot].w = w;
                    tot++;
                }
            }
        }
        tot--;
        sort(edge + 1, edge + tot + 1);         //kruskal生成最小生成树
        for (int i = 1; i <= tot; ++i)
        {
            int u = edge[i].u;
            int v = edge[i].v;
            int fx = find(u);
            int fy = find(v);
            if (fx != fy)
            {
                merge(u, v, 1);
            }
        }
    }
    return 0;
}