因为到某一没有限速的路径速度会有不同的可能，所以直接用 dis[i][j] 表示到第 i 个点速度为 j 时的最短时间，然后跑spfa。
——代码
1 #include <queue> 2 #include <cstdio> 3 #include <cstring> 4 5 using namespace std; 6 7 const int MAXN = 151; 8 int n, m, d, cnt; 9 int head[MAXN], to[MAXN * MAXN], next[MAXN * MAXN], spd[MAXN * MAXN], lon[MAXN * MAXN], pr[MAXN][510], ps[MAXN][510]; 10 double dis[MAXN][510], ans = 12345678; 11 bool vis[MAXN][510]; 12 queue < pair <int, int> > q; 13 pair <int, int> x; 14 15 inline void add(int x, int y, int v, int l) 16 { 17 to[cnt] = y; 18 spd[cnt] = v; 19 lon[cnt] = l; 20 next[cnt] = head[x]; 21 head[x] = cnt++; 22 } 23 24 inline void spfa() 25 { 26 int i, j, u, v, s, p; 27 memset(dis, 70, sizeof(dis)); 28 q.push(make_pair(0, 70)); 29 dis[0][70] = 0; 30 while(!q.empty()) 31 { 32 x = q.front(); 33 q.pop(); 34 u = x.first; 35 s = x.second; 36 vis[u][s] = 0; 37 for(i = head[u]; i != -1; i = next[i]) 38 { 39 v = to[i]; 40 p = spd[i] == 0 ? s : spd[i]; 41 if(dis[v][p] > dis[u][s] + 1.0 * lon[i] / p) 42 { 43 dis[v][p] = dis[u][s] + 1.0 * lon[i] / p; 44 pr[v][p] = u; 45 ps[v][p] = s; 46 if(!vis[v][p]) 47 { 48 vis[v][p] = 1; 49 q.push(make_pair(v, p)); 50 } 51 } 52 } 53 } 54 } 55 56 inline void print(int u, int pos) 57 { 58 if(pr[u][pos] != -1) print(pr[u][pos], ps[u][pos]); 59 printf("%d ", u); 60 } 61 62 int main() 63 { 64 int i, x, y, v, l, pos; 65 scanf("%d %d %d", &n, &m, &d); 66 memset(pr, -1, sizeof(pr)); 67 memset(ps, -1, sizeof(ps)); 68 memset(head, -1, sizeof(head)); 69 for(i = 1; i <= m; i++) 70 { 71 scanf("%d %d %d %d", &x, &y, &v, &l); 72 add(x, y, v, l); 73 } 74 spfa(); 75 for(i = 0; i <= 500; i++) 76 if(ans > dis[d][i]) 77 ans = dis[d][i], pos = i; 78 print(d, pos); 79 return 0; 80 }