一:dijkstra算法时间复杂度,用优先级队列优化的话,O((M+N)logN)求单源最短路径,要求所有边的权值非负.若图中出现权值为负的边,Dijkstra算法就会失效,求出的最短路径就可能是错的. 设road[i][j]表示相邻的i到j的路长U集合存储已经求得的到源点最短路径的节点,S集合表示还没求得的节点dis[i]表示i到源节点(设为0)的最短路径vis[i]=1表示i节点在U集合中 刚开始dis[0]=0,vis[0]=1;dis[i]=maxn,vis[i]=0;for 1 to
链式前向星 在做图论题的时候,偶然碰到了一个数据量很大的题目,用vector的邻接表直接超时,上网查了一下发现这道题数据很大,vector可定会超的,不会指针链表的我找到了链式前向星这个好东西,接下来就由一道裸模板题看看链式前向星怎么写,他的优势又在哪里! 题目链接:POJ 2387 Description Bessie is out in the field and wants to get back to the barn to get as much sleep as possible b
洛谷传送门--分糖果 博客--链式前向星 团队中一道题,数据很大,只能用链式前向星存储,spfa求单源最短路. 可做模板. #include <cstdio> #include <queue> #include <cstring> #include <algorithm> using namespace std; int n, p, c, ans, cnt; long long m; struct node { int to, next; }edge[];
Drainage Ditches Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17065 Accepted Submission(s): 8066 Problem Description Every time it rains on Farmer John's fields, a pond forms over Bessie'
描述 Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a straight line waiting for feed. The cows are standing in the same order as they are numbered, and