johnson最短路

https://www.starrycoding.com/problem/100

关键就在于，解决dijkstra不能处理负权值的问题，先用spfa或者bellman-ford得到所有点到虚点的最短距离，可以称之为势能。然后通过势能对所有边进行处理，处理过后保证了所有边都是正的，再对每个点跑dijkstra就能实现全源带负权的最短路。跑完后再通过势能还原回来。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int M = 3e3 + 5;
const ll inf = 4e18 + 5;int n, m;struct Edge
{ll y, w;bool const operator < (const Edge& a)const{return w == a.w ? y < a.y : w > a.w;//注意此处大于号}
};vector<Edge>g[M];
ll d[M][M];//记录全源最短路
ll h[M];//虚点的单源最短路（势能）
bitset<M>vis;
queue<int>q;
int cnt[M];int spfa(int st)
{q.push(0);vis[0] = 1;while(q.size()){int x = q.front();q.pop();vis[x] = 0;for (auto& [y, w] : g[x]){if (h[y] > h[x] + w){h[y] = h[x] + w;cnt[y]++;if (cnt[y] >= n)return 0;if (!vis[y]){vis[y] = 1;q.push(y);}}}}return 1;
}void dijkstra(int st, ll d[])
{for (int i = 1; i <= n; i++)d[i] = inf;priority_queue<Edge>pq;vis.reset();pq.push({ st,d[st] = 0 });while (pq.size()){int x = pq.top().y; pq.pop();if (vis[x])continue;vis[x] = true;for (auto &[y, w] : g[x]){if (d[y] > d[x] + w){d[y] = d[x] + w;pq.push({ y, d[y] });}}}//还原回来for (int i = 1; i <= n; ++i)d[i] = d[i] - h[st] + h[i];
}void solve()
{cin >> n >> m;for (int i = 1; i <= m; i++){ll x, y, z;cin >> x >> y >> z;g[x].push_back({ y,z });}//创建虚点0，默认0到所有点的距离是0for (int i = 1; i <= n; i++){g[0].push_back({ i,0 });h[i] = inf;}//spfa得出虚点最短路if (!spfa(0))//如果存在负环直接跳出{cout << "starrycoding" << '\n';return;}//将所有边进行改动for (int i = 1; i <= n; i++){for (auto& [y, w] : g[i])//一定要加引用！！w = w + h[i] - h[y];}//每个点跑一遍 dijkstrafor (int i = 1; i <= n; i++)dijkstra(i, d[i]);int q;cin >> q;while (q--){int x, y;cin >> x >> y;//有的点可能被更新过，所以此处用inf/2if (d[x][y] >= inf / 2)cout << "noway" << '\n';elsecout << d[x][y] << '\n';}
}int main()
{std::ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);solve();return 0;
}

其他博主比较好的文章
https://blog.csdn.net/hzf0701/article/details/107697261