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法律研究有向图中的欧拉回路
欧拉路径是一条路径，我们可以通过它精确地访问每条边一次。我们可以多次使用相同的顶点。欧拉回路是一种特殊的欧拉路径。当欧拉路径的起始顶点也与该路径的结束顶点相连时，则称为欧拉回路。
要检查图是否为欧拉图，我们必须检查两个条件：
- 图必须是连通的。
- 每个顶点的入度和出度必须相同。
输入和输出
Input: Adjacency matrix of the graph. 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 0 0 Output: Euler Circuit Found.
算法
traverse(u, visited)
输入：起始节点 u 和已访问节点，用于标记哪个节点已被访问。
输出：遍历所有连接的顶点。
Begin mark u as visited for all vertex v, if it is adjacent with u, do if v is not visited, then traverse(v, visited) done End
isConnected(graph)
输入：图。
输出：如果图是连通的，则返回 True。
Begin define visited array for all vertices u in the graph, do make all nodes unvisited traverse(u, visited) if any unvisited node is still remaining, then return false done return true End
isEulerCircuit(Graph)
输入：给定的图。
输出：找到一个欧拉回路时返回 True。
Begin if isConnected() is false, then return false define list for inward and outward edge count for each node for all vertex i in the graph, do sum := 0 for all vertex j which are connected with i, do inward edges for vertex i increased increase sum done number of outward of vertex i is sum done if inward list and outward list are same, then return true otherwise return false End
示例
#include<iostream>
#include<vector>
#define NODE 5
using namespace std;
int graph[NODE][NODE] = {
{0, 1, 0, 0, 0},
{0, 0, 1, 0, 0},
{0, 0, 0, 1, 1},
{1, 0, 0, 0, 0},
{0, 0, 1, 0, 0}
};
void traverse(int u, bool visited[]) {
visited[u] = true; //mark v as visited
for(int v = 0; v<NODE; v++) {
if(graph[u][v]) {
if(!visited[v])
traverse(v, visited);
}
}
}
bool isConnected() {
bool *vis = new bool[NODE];
//for all vertex u as start point, check whether all nodes are visible or not
for(int u; u < NODE; u++) {
for(int i = 0; i<NODE; i++)
vis[i] = false; //initialize as no node is visited
traverse(u, vis);
for(int i = 0; i<NODE; i++) {
if(!vis[i]) //if there is a node, not visited by traversal, graph is not connected
return false;
}
}
return true;
}
bool isEulerCircuit() {
if(isConnected() == false) { //when graph is not connected
return false;
}
vector<int> inward(NODE, 0), outward(NODE, 0);
for(int i = 0; i<NODE; i++) {
int sum = 0;
for(int j = 0; j<NODE; j++) {
if(graph[i][j]) {
inward[j]++; //increase inward edge for destination vertex
sum++; //how many outward edge
}
}
outward[i] = sum;
}
if(inward == outward) //when number inward edges and outward edges for each node is same
return true;
return false;
}
int main() {
if(isEulerCircuit())
cout << "Euler Circuit Found.";
else
cout << "There is no Euler Circuit.";
}输出
Euler Circuit Found.
更新于： 2020年6月16日
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