4.2 Given a directed graph, design an algorithm to find out whether there is a route between two nodes. 1.圖的存儲(chǔ)使用鄰接矩陣 2.使用鄰接矩陣的過程中,必須給每個(gè)節(jié)點(diǎn)編號(hào)(0,1...) 3.可以寫一個(gè)map函數(shù)給按一定規(guī)則所有節(jié)點(diǎn)編號(hào)(本例未實(shí)現(xiàn)) 4.alg
4.2 Given a directed graph, design an algorithm to find out whether
there is a route between two nodes.
1.圖的存儲(chǔ)使用鄰接矩陣
2.使用鄰接矩陣的過程中,必須給每個(gè)節(jié)點(diǎn)編號(hào)(0,1...)
3.可以寫一個(gè)map函數(shù)給按一定規(guī)則所有節(jié)點(diǎn)編號(hào)(本例未實(shí)現(xiàn))
4.algorithm:通過bfs或dfs遍歷從其中一個(gè)節(jié)點(diǎn)開始遍歷圖,看是否對(duì)可達(dá)另一個(gè)節(jié)點(diǎn)。
import java.util.Queue; import java.util.LinkedList; import java.util.Stack; //vertex in the graph must have serial number //from 0 to n, if graph isn't suitable for this //write map() to map. //directed no-weight graph class Graph1{ private static boolean[][] Matrix; private static int vertexNum; public static void generator(int[][] G, int vNum){ vertexNum = vNum; Matrix = new boolean[vertexNum][vertexNum]; for(int i=0;i < G.length;i++){ Matrix[G[i][0]][G[i][1]] = true; } } public static boolean isRouteDFS(int v1, int v2){ if(v1 < 0 || v2 < 0 || v1 >= Matrix.length || v2 >= Matrix.length) return false; boolean[] isVisited = new boolean[vertexNum]; Stacks = new Stack (); int v = -1; if(v1 != v2){ s.push(v1); isVisited[v1] = true; } else return true; while(!s.empty()){ v = s.peek();//when v's all next node have been invisted, then pop // System.out.println("["+v+"]");// boolean Marked = false; for(int j=0;j < Matrix[v].length;j++) if(Matrix[v][j] && !isVisited[j]) if(j == v2){ return true; } else{ s.push(j); isVisited[j] = true; Marked = true; break; } if(!Marked) s.pop(); } return false; } public static boolean isRouteBFS(int v1, int v2){ if(v1 < 0 || v2 < 0 || v1 >= Matrix.length || v2 >= Matrix.length) return false; boolean[] isVisited = new boolean[vertexNum]; Queue queue = new LinkedList (); int v = -1; queue.offer(v1); isVisited[v1] = true; while(!queue.isEmpty()){ v = queue.poll(); if(v == v2) return true; for(int j = 0;j < Matrix[v].length;j++) if(Matrix[v][j] == true && isVisited[j] == false) queue.offer(j); } return false; } public static void printMatrix(){ for(int i=0;i < Matrix.length;i++){ System.out.print(i+": "); for(int j=0;j < Matrix[i].length;j++) System.out.print((Matrix[i][j] == true ? 1 : 0) + " "); System.out.println(); } } } public class Solution{ public static void main(String[] args){ int[][] G = { {0, 1}, {0, 2}, {1, 3}, {1, 4}, {2, 4}, {4, 0} }; Graph1.generator(G, 5); Graph1.printMatrix(); System.out.println("R :"+Graph1.isRouteBFS(2, 3)); System.out.println("R :"+Graph1.isRouteDFS(2, 3)); } }
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