公共自行车管理 （有些特殊）（dijk+dfs）

https://www.acwing.com/problem/content/description/1497/

思路：
一般这种题型的dijk，我们都可以在dijk中判断出来几个关键字，但这题却不行，因为我们无法确定在哪种情况下更好。所以我们先用一遍dijk来剪枝，找出所有到终点的最短路，然后用dfs暴力搜索所有能到终点的最短路，找到这些最短路中符合题目要求的答案。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

const int N = 510, INF = 0x3f3f3f3f;

int C, n, S, m;
int c[N];
int g[N][N];
int dist[N];
bool st[N];

vector<int> path, ans;
int send = INF, bring = INF;

void dijkstra()
{
    memset(dist, 0x3f, sizeof dist);
    dist[S] = 0;

    for (int i = 0; i < n; i ++ )
    {
        int t = -1;
        for (int j = 0; j <= n; j ++ )
            if (!st[j] && (t == -1 || dist[j] < dist[t]))
                t = j;

        st[t] = true;
        for (int j = 0; j <= n; j ++ )
            dist[j] = min(dist[j], dist[t] + g[t][j]);
    }
}

void dfs(int u, int s, int mins)
{
    if (u)
    {
        s -= (C + 1) / 2 - c[u];
        mins = min(mins, s);
    }

    if (u == S)
    {
        int sd = abs(min(mins, 0));
        int bg = s + sd;

        if (sd < send) ans = path, send = sd, bring = bg;
        else if (sd == send && bg < bring) ans = path, bring = bg;

        return;
    }

    for (int i = 1; i <= n; i ++ )
        if (dist[u] == g[u][i] + dist[i])
        {
            path.push_back(i);
            dfs(i, s, mins);
            path.pop_back();
        }
}

int main()
{
    cin >> C >> n >> S >> m;
    for (int i = 1; i <= n; i ++ ) cin >> c[i];

    memset(g, 0x3f, sizeof g);
    for (int i = 0; i < m; i ++ )
    {
        int x, y, z;
        cin >> x >> y >> z;
        g[x][y] = g[y][x] = min(g[x][y], z);
    }

    dijkstra();

    path.push_back(0);   
    dfs(0, 0, 0);

    cout << send << ' ' << 0;
    for (int i = 1; i < ans.size(); i ++ )
        cout << "->" << ans[i];
    cout << " " << bring << endl;

    return 0;
}