Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 36860   Accepted: 13505 Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way p…

点击打开链接 Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 25582   Accepted: 9186 Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one…

 POJ 3259 Wormholes Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu   Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way…

一个人的旅行 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 23416    Accepted Submission(s): 8154 Problem Description 虽然草儿是个路痴(就是在杭电待了一年多,居然还会在校园里迷路的人,汗~),但是草儿仍然很喜欢旅行,因为在旅途中 会遇见很多人(白马王子,^0^),很多事,还能丰富…

最短路 Time Limit : 5000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 82   Accepted Submission(s) : 51 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…

//差分约束 >=求最长路径 <=求最短路径 结果都一样//spfa#include<stdio.h> #include<string.h> #include<limits.h> #include<queue> using namespace std; #define N 1010 #define M 1010*1010//注意边和点集的数组大小 struct edge { int to,value,next; }; struct edge ed…

Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 49962   Accepted: 18421 Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way p…

POJ3259 :Wormholes 时间限制:2000MS 内存限制:65536KByte 64位IO格式:%I64d & %I64u 描述 While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destin…

Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Eac…

题目链接:http://poj.org/problem?id=3259 Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 55082   Accepted: 20543 Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is ver…

J - Wormholes Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Description In the year 2163, wormholes were discovered. A wormhole is a subspace tunnel through space and time connecting two star systems. Wormholes…

版权声明:本文作者靖心,靖空间地址:http://blog.csdn.net/kenden23/,未经本作者同意不得转载. https://blog.csdn.net/kenden23/article/details/37738183 本题事实上也能够使用SPFA算法来求解的,只是就一个关键点,就是当某个顶点入列的次数超过全部顶点的总数的时候,就能够推断是有负环出现了. SPFA原来也是能够处理负环的. 只是SPFA这样的处理负环的方法自然比一般的Bellman Ford算法要慢点了. #inc…

一种路是双向的,路的长度是正值:另一种路是单向的,路的长度是负值:  如果有负环输出YES:否则输出NO:不同的路可能有相同的起点和终点:必须用邻接表 While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its de…

题目链接:http://poj.org/problem?id=3259 题目大意是给你n个点,m条双向边,w条负权单向边.问你是否有负环(虫洞). 这个就是spfa判负环的模版题,中间的cnt数组就是记录这个点松弛进队的次数,次数超过点的个数的话,就说明存在负环使其不断松弛. #include <iostream> #include <cstdio> #include <cstring> #include <queue> using namespace st…

题意:有向图判负环. 解题关键:spfa算法+hash判负圈. spfa判断负环:若一个点入队次数大于节点数,则存在负环.  两点间如果有最短路,那么每个结点最多经过一次,这条路不超过$n-1$条边.” 如果一个结点经过了两次,那么我们走了一个圈.如果这个圈的权为正,显然不划算:如果是负圈,那么最短路不存在:如果是零圈,去掉不影响最优值. 也就是说,每个点最多入队$n-1$次,可以想象一下,左边$n-1$个节点全部指向右边一个节点,遍历的顺序恰好与边权顺序相反. 负圈是指圈上的总和小于0 实际只…

问题描述 John在他的农场中闲逛时发现了许多虫洞.虫洞可以看作一条十分奇特的有向边,并可以使你返回到过去的一个时刻(相对你进入虫洞之前).John的每个农场有M条小路(无向边)连接着N (从1..N标号)块地,并有W个虫洞.其中1<=N<=500,1<=M<=2500,1<=W<=200. 现在John想借助这些虫洞来回到过去(出发时刻之前),请你告诉他能办到吗. John将向你提供F(1<=F<=5)个农场的地图.没有小路会耗费你超过10000秒的时间,…

题意 : 给出 N 个点,以及 M 条双向路,每一条路的权值代表你在这条路上到达终点需要那么时间,接下来给出 W 个虫洞,虫洞给出的形式为 A B C 代表能将你从 A 送到 B 点,并且回到 C 个时间点之前,也就是时光倒流了 C 个时间并且此时你在 B 点,现在问你是否能够在图上的这些点中走,使得在某一个点刚好碰到之前的自己 分析 : 冷静分析一下,只要是有负权回路的某一条边属于虫洞的,那么肯定是能在这个环上一直绕,直到遇到之前的自己,如果将虫洞看作一条负权边的话,那么问题就变成了只要存在负…

-----------------------------最优化问题------------------------------------- ----------------------常规动态规划  SOJ1162 I-Keyboard  SOJ1685 Chopsticks SOJ1679 Gangsters SOJ2096 Maximum Submatrix  SOJ2111 littleken bg SOJ2142 Cow Exhibition  SOJ2505 The County…

图论 最短路:dijkstra   spfa   floyd 最小生成树:kruskal 连通性:bfs/dfs    tarjan(强连通分量) 其它:拓扑排序    LCA 齿轮: 图的dfs树只有返祖边没有横向边 树边确定了基础的转动比,非树边判定转动比是否可行 Lca Dijkstra Pair第一个存的是源点到这个点当前的最短路,第二个存的是这个点的编号 预处理 把1压进去,1到其他的dis都设置为正无穷 不断地从堆顶弹出元素,直到弹空 对其中的元素进行处理 注意如果已经处理过就不用处…

问题描述 给定一个n个顶点,m条边的有向图(其中某些边权可能为负,但保证没有负环).请你计算从1号点到其他点的最短路(顶点从1到n编号). 输入格式 第一行两个整数n, m. 接下来的m行,每行有三个整数u, v, l,表示u到v有一条长度为l的边. 输出格式 共n-1行,第i行表示1号点到i+1号点的最短路. 样例输入 3 3 1 2 -1 2 3 -1 3 1 2 样例输出 -1 -2 由于有负边,所以Dijkstra算法无法使用,而Floyd算法的效率太低,容易超时,所以使用Bellman…

题目链接:558 - Wormholes 题目大意:给出n和m,表示有n个点,然后给出m条边,然后判断给出的有向图中是否存在负环. 解题思路:利用Bellman Ford算法,若进行第n次松弛时,还能更新点的权值,则说明有负环的存在. #include <stdio.h> #include <string.h> #define min(a,b) (a)<(b)?(a):(b) const int N = 10005; const int INF = 0x3f3f3f3f; i…

http://poj.org/problem?id=3259 Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFOR…

Wormholes DescriptionWhile exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the worm…

Wormholes Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 42366 Accepted: 15560 传送门 Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way p…

Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 36729   Accepted: 13444 Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way p…

watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvdTAxMzQ4NzA1MQ==/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast" width="900" height="500" alt=""> watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvd…

Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Eac…

题目电波   3259 Wormholes #include<iostream> #include<cstring> #include<algorithm> #include<queue> #include<stdio.h> using namespace std; #define maxn 100010 #define inf 0x3f3f3f3f struct ac{ int to,va; ac(){} ac(int a,int b){ to…

数据结构与算法--最短路径之Bellman算法.SPFA算法 除了Floyd算法,另外一个使用广泛且可以处理负权边的是Bellman-Ford算法. Bellman-Ford算法 假设某个图有V个顶点E条边. 该算法主要流程是: 初始化.到起点s的距离distTo[s]设置为0,其余顶点的dist[]设置为正无穷: 以任意次序放松图中的所有E条边,重复V轮: V轮放松结束后,判断是否存在负权回路.如果存在,最短路径没有意义. 根据流程可以给出代码,如下 package Chap7; import…

http://poj.org/problem?id=3259 Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 37356   Accepted: 13734 Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very pec…