題意

給你一個n個點，m條邊的DAG圖，邊為雙向邊，沒有重邊。現在最多保留k條邊，怎麼使得好點個數最多。 好點定義為：在原圖中1到該點距離和只保留某一些邊後的圖中1到該點距離不變的點。 先輸出保留邊的個數，然後輸出這些保留的邊的編號（1~m）。

思路

堆優化dijkstra中，更新前 k 個點用到的邊就是答案。 資料比較大，注意開long long

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 3e5+5;
typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f
 
;

struct Edge
{
    int from, to; LL dist;       //起點,終點,距離
    Edge(int from, int to, LL dist):from(from), to(to), dist(dist) {}
};

struct Dijkstra
{
    int n, m;                 //結點數,邊數(包括反向弧)
    vector<Edge> edges;       //邊表。edges[e]和edges[e^1]互為反向弧
    vector<int> G[MAXN];      //鄰接表，G[i][j]表示結點i的第j條邊在edges陣列中的序號
 

    int vis[MAXN];            //標記陣列
    LL d[MAXN];              //s到各個點的最短路
    int p[MAXN];              //上一條弧

    void init(int n)
    {
        this->n = n;
        edges.clear();
        for (int i = 0; i <= n; i++) G[i].clear();
    }

    void AddEdge(int from, int to, int dist)
    {
        edges.
 
emplace_back(from, to, dist);
        m = edges.size();
        G[from].push_back(m - 1);
    }

    struct HeapNode
    {
        int from; LL dist;
        bool operator < (const HeapNode& rhs) const
        {
            return rhs.dist < dist;
        }
        HeapNode(int u, LL w): from(u), dist(w) {}
    };

    void dijkstra(int s, vector<int>& ans, int k)
    {
        priority_queue<HeapNode> Q;
        for (int i = 0; i <= n; i++) d[i] = INF;
        memset(vis, 0, sizeof(vis));
        d[s] = 0;
        Q.push(HeapNode(s, 0));
        while (!Q.empty() && ans.size() <= k)
        {
            HeapNode x = Q.top(); Q.pop();
            int u = x.from;
            if (vis[u]) continue;
            vis[u] = true;
            ans.push_back(p[u]/2);
            for (int i = 0; i < G[u].size(); i++)
            {
                Edge& e = edges[G[u][i]];
                if (d[e.to] > d[u] + e.dist)
                {
                    d[e.to] = d[u] + e.dist;
                    p[e.to] = G[u][i];
                    Q.push(HeapNode(e.to, d[e.to]));
                }
            }
        }
    }
}gao;

int n, m, k;
int vis[MAXN];
int main()
{
    scanf("%d%d%d", &n, &m, &k);
    memset(vis, 0, sizeof(vis));
    gao.init(n);
    while (m--)
    {
        int x, y, w; scanf("%d%d%d", &x, &y, &w);
        gao.AddEdge(x, y, w);
        gao.AddEdge(y, x, w);
    }
    vector<int> ans;
    gao.dijkstra(1, ans, k);
    printf("%d\n", ans.size()-1);
    for (int i = 1; i < ans.size(); i++)
        printf("%d%c", ans[i]+1, i==ans.size()-1?'\n':' ');
    return 0;
}