[AcWing 1135] 新年好

最短路 + DFS 枚举顺序

最短路用堆优化 dijkstra 复杂度 \(O(m \cdot log(n)) = 10^5 \cdot log(5 \times 10^4) \approx 1.56 \times 10^6\)

DFS 枚举复杂度 \(O(n!) = 5! = 120\)

两个复杂度是相加的关系

---

点击查看代码

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int,int> PII;

const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;

int n, m;
int h[N], e[N], ne[N], w[N], idx;
int d[6][N]; // 第一维是 0 ~ 5，代表的是人，而不是车站
int stop[6];
bool st[N];

void add(int a, int b, int c)
{
    e[idx] = b;
    w[idx] = c;
    ne[idx] = h[a];
    h[a] = idx ++;
}

void dijkstra(int sp, int d[])
{
    memset(d, 0x3f, sizeof st);
    memset(st, false, sizeof st);
    priority_queue<PII, vector<PII>, greater<PII>> heap;
    heap.push({0, sp});
    d[sp] = 0;
    while (heap.size()) {
        auto t = heap.top();
        heap.pop();
        auto v = t.second;
        if (st[v])
            continue;
        st[v] = true;
        for (int i = h[v]; i != -1; i = ne[i]) {
            int j = e[i];
            if (d[j] > d[v] + w[i]) {
                d[j] = d[v] + w[i];
                heap.push({d[j], j});
            }
        }
    }
}

int dfs(int u, int start, int dist)
{
    if (u > 5)
        return dist;
    int res = INF;
    for (int i = 1; i <= 5; i ++)
        if (!st[i]) {
            int next = stop[i];
            st[i] = true;
            res = min(res, dfs(u + 1, i, dist + d[start][next]));
            st[i] = false;
        }
    return res;
}

void solve()
{
    cin >> n >> m;
    stop[0] = 1;
    for (int i = 1; i <= 5; i ++)
        cin >> stop[i];
    memset(h, -1, sizeof h);
    for (int i = 0; i < m; i ++) {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
        add(b, a, c);
    }
    // 从 stop[i] 出发的最短路
    for (int i = 0; i < 6; i ++)
        dijkstra(stop[i], d[i]);
    // DFS 枚举选择顺序
    memset(st, false, sizeof st);
    cout << dfs(1, 0, 0) << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    solve();

    return 0;
}

---

1. 先用 \(6\) 次 \(dijkstra\) 求出从 \(6\) 个点（佳佳和五个亲戚）出发到其他点的最短距离，用 \(DFS\) 枚举访问亲戚的顺序，由于前面已经用 \(dijkstra\) 求出了最短距离，算距离时直接以 \(O(1)\) 的复杂度从 \(d\) 数组中得到