分析：出題人喪心病狂卡spfa......只能用dijkstar+堆優化.

主要的難點是字典序的處理上，一個想法是在做最短路的時候處理，邊松弛邊記錄，比個大小記錄最佳答案.具體的思路大概和最短路計數差不多，當遇到d[u] + w[i] == d[v]是，說明到d[v]有兩條最短路了，更新一下答案。

但是這樣效率太低了，每一次松弛都要更新一次，能不能一次性更新完呢？可以的.我們在跑完最短路後把滿足d[u] + w[i] == d[v]的點v加入u的鄰接表中，排個序，然後dfs一遍就好了.

#include <cstdio>
#include <cstring>
#include 
 
<iostream>
#include <algorithm>
#include <string>
#include <cmath>
#include <queue>
#include <vector>

using namespace std;

const long long inf = 1e15;

struct node
{
    long long len;
    int x;
};

bool operator < (node a, node b)
{
    return a.len > b.len;
}

 
int n, m,b[200010],head[200010],to[400010],nextt[400010],tot = 1,w[400010],vis[200010];
int u[400010], v[400010], z[400010],ans[200010];
long long d[200010];
vector <int> a[200010];

void add(int x, int y, int z)
{
    w[tot] = z;
    to[tot] = y;
    nextt[tot] = head[x];
    head[x] = tot++;
}

void dijkstar()
{
    d[1] = 0;
    priority_queue 
 
<node> q;
    node t;
    t.len = 0;
    t.x = 1;
    q.push(t);
    while (!q.empty())
    {
        while (!q.empty() && vis[q.top().x])
            q.pop();
        if (q.empty())
            break;
        node u = q.top();
        q.pop();
        long long len = u.len;
        int x = u.x;
        vis[x] = 1;
        for (int i = head[x]; i;i = nextt[i])
        {
            int v = to[i];
            if (len + w[i] < d[v])
            {
                d[v] = len + w[i];
                node temp;
                temp.len = d[v];
                temp.x = v;
                q.push(temp);
            }
        }
    }
}

bool cmp(int x, int y)
{
    return b[x] < b[y];
}

void dfs(int depth, int x)
{
    if (vis[x])
        return;
    vis[x] = 1;
    ans[depth] = x;
    if (x == n)
    {
        for (int i = 1; i <= depth; i++)
            printf("%d ", b[ans[i]]);
        printf("\n");
        return;
    }
    for (int i = 0; i < a[x].size(); i++)
        dfs(depth + 1, a[x][i]);
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%d", &b[i]);
    for (int i = 1; i <= m; i++)
    {
        scanf("%d%d%d", &u[i], &v[i], &z[i]);
        add(u[i], v[i], z[i]);
    }
    for (int i = 1; i <= n; i++)
        d[i] = inf;
    dijkstar();
    printf("%lld\n", d[n]);
    for (int i = 1; i <= m; i++)
        if (d[u[i]] + z[i] == d[v[i]])
            a[u[i]].push_back(v[i]);
    for (int i = 1; i <= n; i++)
        sort(a[i].begin(), a[i].end(), cmp);
    memset(vis, 0, sizeof(vis));
    dfs(1, 1);

    return 0;
}

常州模擬賽d5t1 journalist