描述 一个连通图采用邻接表作为存储结构.设计一个算法,判断无向图中任意给定的两点是否存在一条长度为k的简单路径. 输入 多组数据,每组m+3数据行.第一行有两个数字n,m和k,代表有n个顶点,m条边和长度k.第二行有n个字符,代表n个顶点的编号.第三行到第m+2行每行有两个字符h和p,代表边依附的两个顶点.每条边的长度为1.第m+3行有两个字符d和f,代表需要判断的两个字符. 输出 每组数据输出一行.若存在路径输出“YES”,反之输出“NO”. 输入样例 1 3 2 2 abc ab bc ac
Ramesses knows a lot about problems involving trees (undirected connected graphs without cycles)! He created a new useful tree decomposition, but he does not know how to construct it, so he asked you for help! The decomposition is the splitting the e
<script> //图的构建 function vnode() { this.visited=0; this.vertex=0; this.arcs=new Array(); } function G() { this.adjlist=new Array(); } function creategraph() { var a=[[2,3,6],[1,5,4],[4,6,1],[2,3,5],[4,2,6],[3,5,1]]; var g=new G(); for (i=0;i<6;i+
import java.sql.*; public class JDBCExample { static final String JDBC_DRIVER = "com.mysql.jdbc.Driver"; static final String DB_URL = "jdbc:mysql://localhost:3306/gbooks"; static final String USER = "root"; static final Strin
03-树1. List Leaves (25) Given a tree, you are supposed to list all the leaves in the order of top down, and left to right. Input Specification: Each input file contains one test case. For each case, the first line gives a positive integer N (<=10) wh
Plan Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 686 Accepted Submission(s): 234 Problem Description One day, Resty comes to an incredible world to seek Eve -- The origin of life. Lilith
判断从顶点u到v是否有路径 void ExistPath(AdjGraph* G, int u, int v, bool& has) { int w; ArcNode* p; visit[u] = 1; if (u == v) { has = true; return; } p = G->adjlist[u].firstarc; while (p != NULL) { w = p->adjvex; if (visit[w] == 0) ExistPath(G, w, v, has);