dijkstra

最短路径问题 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 43967 Accepted Submission(s): 12599 Problem Description 给你n个点，m条无向边，每条边都有长度d和花费p，给你起点s终点t，要求输出起点到终点的最短距离及其花费，如果最短距离有多条路线，则输出花费最少的。 Input 输入n,m，点的编号是1~n,然后是m行，每行4个数 a,b,d,p，表示a和b之间有一条边，且其长度为d，花费为p。最后一行是两个数 s,t;起点s，终点。n和m为0时输入结束。 (1<n<=1000, 0<m<100000, s != t) Output 输出 一行有两个数， 最短距离及其花费。 Sample Input 3 2 1 2 5 6 2 3 4 5 1 3 0 0 Sample Output 9 11 Source 浙大计算机研究生复试上机考试-2010年 这道题用floyd算法会超时 两个权值的Dijkstra。。 #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include

最短路 Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 109250 Accepted Submission(s): 46973 Problem Description 在每年的校赛里，所有进入决赛的同学都会获得一件很漂亮的t-shirt。但是每当我们的工作人员把上百件的衣服从商店运回到赛场的时候，却是非常累的！所以现在他们想要寻找最短的从商店到赛场的路线，你可以帮助他们吗？ Input 输入包括多组数据。每组数据第一行是两个整数N、M（N<=100，M<=10000），N表示成都的大街上有几个路口，标号为1的路口是商店所在地，标号为N的路口是赛场所在地，M则表示在成都有几条路。N=M=0表示输入结束。接下来M行，每行包括3个整数A，B，C（1<=A,B<=N,1<=C<=1000）,表示在路口A与路口B之间有一条路，我们的工作人员需要C分钟的时间走过这条路。 输入保证至少存在1条商店到赛场的路线。 Output 对于每组输入，输出一行，表示工作人员从商店走到赛场的最短时间 Sample Input 2 1 1 2 3 3 3 1 2 5 2 3 5 3 1 2 0 0 Sample Output 3 2 Source

I was asked this question in an interview, but I couldn't come up with any decent solution. So, I told them the naive approach of finding all the cycles then picking the cycle with the least length. I'm curious to know what is an efficient solution to this problem. You can easily modify Floyd-Warshall algorithm . (If you're not familiar with graph theory at all, I suggest checking it out, e.g. getting a copy of Introduction to Algorithms ). Traditionally, you start path[i][i] = 0 for each i . But you can instead start from path[i][i] = INFINITY . It won't affect algorithm itself, as those

该文章可能存在硬伤与不妥， 不能 作为教程阅读。 Dij作为单源最短路算法，需要先确定一个起点。Dij的函数主体为维护每个节点的dis和vis两个变量。dis表示该点距离起点的最短路权值和，vis存储该点是否被访问过。 设点数为n，无向边数为m，则要进行n次循环，每次循环中找出一个dis最小且vis不为真的点进行松弛操作。 这里抄一段百科：松弛操作是指对于每个顶点v∈V，都设置一个属性d[v]，用来描述从源点s到v的最短路径上权值的上界，称为最短路径估计(shortest-path estimate)。 然后就是优化： 　　(1)判断最小值的循环可用优先队列优化 　　(2)使用优先队列后可以直接判断堆中是否剩余元素来循环 　　(3)使用链表存储图，松弛操作只遍历相关的边 模板： #include <bits/stdc++.h> #define maxn 10010 #define maxm 10010 using namespace std; struct edge{ int to, val, next; }; edge g[maxm]; int first[maxn], edge_cnt = 0; struct node{ int pos, dis; }; priority_queue<node> q; bool operator<(const node a, const

#include <stdio.h> #include <stdlib.h> #include <stdbool.h> #include <string.h> #define MaxSize 100 #define INF 1000001 typedef struct EageTable { int node,len; struct EageTable *next; }EageTable; typedef struct HeadTable { int node; EageTable *first; }HeadTable,HeadList[MaxSize]; typedef struct { HeadList List; int nodeNum,arcNum; }Graph; int dis[MaxSize];//记录距离 bool vis[MaxSize] = {false};//记录是否访问过 void InitGraph(Graph *G); void CreateGraph(Graph *G,int flag,int a,int b,int len); void Dijkstra(Graph *G,int start); /*初始化图*/ void InitGraph(Graph *G) { int i = 0; for(;i < MaxSize;i++) { G->List

Can we use Dijkstra's algorithm with negative weights? STOP! Before you think "lol nub you can just endlessly hop between two points and get an infinitely cheap path", I'm more thinking of one-way paths. An application for this would be a mountainous terrain with points on it. Obviously going from high to low doesn't take energy, in fact, it generates energy (thus a negative path weight)! But going back again just wouldn't work that way, unless you are Chuck Norris. I was thinking of incrementing the weight of all points until they are non-negative, but I'm not sure whether that will work. D

Both can be used to find the shortest path from single source. BFS runs in O(E+V) , while Dijkstra's runs in O((V+E)*log(V)) . Also, I've seen Dijkstra used a lot like in routing protocols. Thus, why use Dijkstra's algorithm if BFS can do the same thing faster? Dijkstra allows assigning distances other than 1 for each step. For example, in routing the distances (or weights) could be assigned by speed, cost, preference, etc. The algorithm then gives you the shortest path from your source to every node in the traversed graph. Meanwhile BFS basically just expands the search by one “step” (link,