[AcWing 920] 最优乘车

按照换乘次数建图，求单源最短路

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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[N];
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)
{
    memset(d, 0x3f, sizeof d);
    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] + 1) {
                d[j] = d[v] + 1;
                heap.push({d[j], j});
            }
        }
    }
}

void solve()
{
    cin >> m >> n;
    memset(h, -1, sizeof h);
    string line;
    // 换行
    getline(cin, line);
    for (int i = 0; i < m; i ++) {
        getline(cin, line);
        stringstream ssin(line);
        vector<int> stop;
        int p = 0;
        while (ssin >> p)
            stop.push_back(p);
        for (int i = 0; i < stop.size(); i ++)
            for (int j = i + 1; j < stop.size(); j ++)
                add(stop[i], stop[j], 1);
    }
    dijkstra(1);
    if (d[n] == INF)
        cout << "NO" << endl;
    else
        cout << d[n] - 1 << endl;
}

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

    solve();

    return 0;
}

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1. 将与起点在同一线路的车站距离置为 $1$，需要换成一次的车站距离置为 $2$，$\cdots$，求从起点到终点的最短路，最后把结果减 $1$（因为和起点在同一线路的实际不需要换乘）