(Prim堆优化)滑雪与时间胶囊

滑雪与时间胶囊

\(本题可作为\ Prim堆优化\ 求最小生成树的模板题\)

\(由于受高度的限制,所以队列中存的不止要有边权,还得有点的高度,\\如果只存边权,就可能先把高度小的入队,最后发现走不通,又得回到高度大的点\\这个时候统计的ans就不对了.\)

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define ull unsigned long long
#define pb push_back
#define PII pair<int, int>
#define VIT vector<int>
#define x first
#define y second
#define inf 0x3f3f3f3f
const int N = 1e5 + 10, M = 2e6 + 10;
int h[N];
bool st[N];
vector<PII> e[N];
int dist[N];
int n, m;

struct qnode {
    int v, h, dis; 
    qnode(int _v = 0, int _h = 0, int _dis = 0) : v(_v), h(_h), dis(_dis){}
    bool operator < (const qnode &t) const {
        return h == t.h ? dis > t.dis : h < t.h;
    }
};

void prim() {
    ll ans = 0;
    int cnt = 0;
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    priority_queue<qnode> q;
    q.push({1, h[1], 0});
    while(q.size()) {
        qnode t = q.top();
        q.pop();
        int ver = t.v;
        if (st[ver]) continue;
        st[ver] = true;
        ans += dist[ver];
        ++cnt;
        for (int i = 0; i < e[ver].size(); ++i) {
            int j = e[ver][i].y, w = e[ver][i].x;
            if (!st[j] && dist[j] > w) {
                dist[j] = w;
                q.push({j, h[j], w});
            }
        }
    }
    cout << cnt << " " << ans << '\n';    
}

int main() {
    IO;
    //freopen("in.txt", "r", stdin);
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) cin >> h[i];
    while (m--) {
        int a, b, w;
        cin >> a >> b >> w;
        if (h[a] < h[b]) swap(a, b);
        e[a].pb({w, b});
        if (h[a] == h[b]) e[b].pb({w, a});
    }
    prim();
    return 0;
}