dijkstra

单源最短路算法Dijkstra 思路：每次寻找距离原点最近的点，通过该点松弛其他的点。 变量 m为边数，n为点数，dis[i]表示从1到点i的最短路径，book[i]表示点1到点i的路程是否最短，e为邻接矩阵，min和u之后才用到 int m, n, dis[N], book[N], e[N][N] min, u; 初始化+读入 读入n,m不再赘述 scanf("%d %d", &n, &m); 一开始所有点都初始化为INF表示互不相连，i==j时自己连向自己，路程为0 //初始化点i到j的距离为INF 自己到自己为0 for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) { if (i == j) e[i][j] = 0; else e[i][j] = INF; } 读入m条边 将数组e[a][b]的值赋值为权重c //读入 for (int i = 1; i <= m; i++) { int a, b, c; scanf("%d%d%d", &a, &b, &c); e[a][b] = c; } 因为e表示i到j的路程，dis表示从1到i的路程。 所以dis[i]初始化时等于e[1][i] 就是 从1到i的路程 (dis[i]) 等于 从1到 i (e[1][i]) 的路程 翻译成人话就是完全一样的东西

Given a directed, connected graph with only positive edge weights, are there faster algorithms for finding the shortest path between two vertices, than Dijkstra using a fibonacci heap? Wikipedia says, that Dijkstra is in O(|E| + |V| * log(|V|)) (using a fibonacci heap). I'm not looking for optimizations that, for example, half the execution time, but rather algorithms that are in a different time complexity (like going from O(n * log n) to O(n)). Further, I would like to know your opinion on the following approach: Determine the GCD of all edge weights. Transform the graph into a graph with

I am reading up on Dijkstra's algorithm and the Floyd-Warshall algorithm. I understand that Dijkstra's finds the optimal route from one node to all other nodes and Floyd-Warshall finds the optimal route for all node pairings. My question is would Dijkstra's algorithm be more efficient than Floyd's if I run it on every single node in order to find the optimal route between all pairings. Dijkstra's runtime is O(E + VlogV) where Floyd's is O(V 3 ). If Dijkstra's fails, what would its runtime be in this case? Thanks! As others have pointed out, Floyd-Warshall runs in time O(n 3 ) and running a

import random as rd def Dis_generate(nums, n_min, n_max): return [[int((n_max-n_min) * rd.random() + n_min) if i != j else 0 for j in range(nums)] for i in range(nums)] def Dijkstra(Dis): n = len(Dis) R = [] Dis_R = [] for start in range(n): #使用状态、路线、距离 states = [0 if i != start else 1 for i in range(n)] r = ["" if i != start else str(start) for i in range(n)] dis = [Dis[start][i] for i in range(n)] while sum(states) < n: S1 = [i for i in range(n) if states[i] == 1] S2 = [i for i in range(n) if states[i] == 0] temp = [[dis[i] + Dis[i][j] for j in S2] for i in S1] mid, end = S1[0], S2[0] for i

先占个坑：开启算法笔记 迪杰斯特拉( Dijkstra )算法是典型的最短路径算法，计算一个节点到其他节点的最短路径。 它的主要特点是以起始点为中心向外层扩展(广度优先搜索思想)，直到扩展到终点为止。 来源： https://blog.csdn.net/u014426939/article/details/100084449

What is the exact difference between Dijkstra's and Prim's algorithms? I know Prim's will give a MST but the tree generated by Dijkstra will also be a MST. Then what is the exact difference? templatetypedef Prim's algorithm constructs a minimum spanning tree for the graph, which is a tree that connects all nodes in the graph and has the least total cost among all trees that connect all the nodes. However, the length of a path between any two nodes in the MST might not be the shortest path between those two nodes in the original graph. MSTs are useful, for example, if you wanted to physically

The question is simple. How can I struct my Graph in SWI prolog to implement the Dijkstra's algorithm? I have found this but it's too slow for my job. CapelliC That implementation isn't so bad: ?- time(dijkstra(penzance, Ss)). % 3,778 inferences, 0,003 CPU in 0,003 seconds (99% CPU, 1102647 Lips) Ss = [s(aberdeen, 682, [penzance, exeter, bristol, birmingham, manchester, carlisle, edinburgh|...]), s(aberystwyth, 352, [penzance, exeter, bristol, swansea, aberystwyth]), s(birmingham, 274, [penzance, exeter, bristol, birmingham]), s(brighton, 287, [penzance, exeter, portsmouth, brighton]), s

至少得知道floyd的最初形式,dp[k][i][j]，只经过前k个点，i到j的最短路。 所以题目就是问你，任意两点间的最短路，经过的最大编号的点最小是多少，还不含两个端点。 魔改一下dijkstra就行了，每个点跑一次。先比较长度，长度相同比较下最大编号谁的更小。每个点存的经过最大编号并不包含这个点本身。记得特判下起点终点直连的情况。 1 #include <cstdio> 2 #include <queue> 3 using namespace std; 4 typedef long long ll; 5 const int mo = 998244353,MAXN = 1010,MAXM = 4010; 6 const ll inf = 1e14; 7 struct pot 8 { 9 int x,maxn; 10 ll dis; 11 pot (int _x = 0,ll _dis = 0,int _maxn = 0) : x(_x),dis(_dis),maxn(_maxn) {} 12 friend bool operator < (pot a,pot b) 13 { 14 if (a.dis != b.dis) 15 return a.dis > b.dis; 16 return a.maxn > b.maxn; 17 } 18 }; 19 struct dat 20

复杂度O(mlogn) 输入起点s,可以得到从起点到各点的最短路距离数组dis[i] 过程: 1.初始化:清空标记数组,初始化距离数组设为inf,起点距离设为0,开优先队列,搜索起点 2.搜索:取出队首并pop,如果队首节点u的当前最短路比u的原先的最短路大则跳过,否则遍历u的邻接点如果v没有被访问过且u的最短路加上当前邻接边边权小于原先v的最短路,则更新v的最短路且搜索v 1 void dijkstra(int s){ 2 memset(vis,0,sizeof vis); 3 memset(dis,inf,sizeof dis); 4 dis[s]=0; 5 priority_queue<node> q; 6 node a(s,dis[s]); 7 q.push(a); 8 while(q.size()){ 9 node x=q.top();q.pop(); 10 int u=x.pos; 11 if(x.val>dis[u])continue; 12 for(int i=head[u];~i;i=e[i].nex){ 13 int v=e[i].to; 14 if(!vis[v]&&(dis[v]>e[i].w+dis[u])){ 15 dis[v]=e[i].w+dis[u]; 16 a.pos=v,a.val=dis[v]; 17 q.push(a); 18 } 19 }