[CF545E] Paths and Trees - 最短路

Description

给定一张带正权的无向图和一个源点，求边权和最小的最短路径树。

Solution

跑最短路的时候，转移时尽量让当前边的边权最小

记录一下前驱，最后连出来就是答案

#include <bits/stdc++.h>
using namespace std;

#define int long long
const int N = 1e6 + 5;

struct Edge
{
    int u, v, w, id;
};

struct Node
{
    int d;
    int p;
    bool operator<(const Node &rhs) const
    {
        return d > rhs.d;
    }
};

int n, m;
vector<int> g[N]; // 存的是边的 id
vector<Edge> edges;

int vis[N], dis[N], pre[N];

signed main()
{
    ios::sync_with_stdio(false);
    cin >> n >> m;
    for (int i = 1; i <= m; i++)
    {
        int u, v, w;
        cin >> u >> v >> w;
        g[u].push_back(edges.size());
        edges.push_back({u, v, w, i});
        g[v].push_back(edges.size());
        edges.push_back({v, u, w, i});
    }
    int s;
    cin >> s;
    priority_queue<Node> que;
    memset(dis, 0x3f, sizeof dis);
    dis[s] = 0;
    que.push({0, s});
    while (que.size())
    {
        auto [d, p] = que.top();
        que.pop();
        if (vis[p])
            continue;
        vis[p] = 1;
        for (auto i : g[p])
        {
            auto [_, q, w, id] = edges[i];
            if (dis[q] > dis[p] + w)
            {
                dis[q] = dis[p] + w;
                pre[q] = i;
                que.push({dis[q], q});
            }
            else if (dis[q] == dis[p] + w && w < edges[pre[q]].w)
            {
                pre[q] = i;
            }
        }
    }

    vector<int> ans;
    int ansval = 0;
    for (int i = 1; i <= n; i++)
    {
        if (i == s)
            continue;
        ans.push_back(edges[pre[i]].id);
        ansval += edges[pre[i]].w;
    }

    sort(ans.begin(), ans.end());

    cout << ansval << endl;
    for (auto i : ans)
        cout << i << " ";
}