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法學C++程式：檢查無向圖是否包含尤拉路徑
尤拉路徑是一條路徑，我們可以精確地訪問每個節點一次。我們可以多次使用相同的邊。歐拉回路是尤拉路徑的一種特殊型別。當尤拉路徑的起始頂點也與該路徑的結束頂點連線時。
要檢測尤拉路徑，我們必須遵循以下條件：
- 該圖必須是連通的。
- 現在，當無向圖的任何頂點都沒有奇數度時，它就是一個歐拉回路，它也是一條尤拉路徑。
- 當恰好有兩個頂點具有奇數度時，它是一條尤拉路徑。
輸入
輸出
這兩個圖都有尤拉路徑。
演算法
traverse(u, visited)
輸入：起始節點u和已訪問節點，用於標記已訪問的節點。
輸出：遍歷所有連線的頂點。
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)
輸入：圖。
輸出：如果圖是連通的，則返回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
isEulerian(Graph)
輸入：給定的圖。
輸出：當存在歐拉回路或路徑時返回1，當不存在尤拉路徑時返回0。
Begin if isConnected() is false, then return false define list of degree for each node oddDegree := 0 for all vertex i in the graph, do for all vertex j which are connected with i, do increase degree done if degree of vertex i is odd, then increase oddDegree done if oddDegree > 0, then return 0 else return 1 End
示例程式碼
#include<iostream>
#include<vector>
#define NODE 5
using namespace std;
int graph[NODE][NODE] = {{0, 1, 1, 1, 0},
{1, 0, 1, 0, 0},
{1, 1, 0, 0, 0},
{1, 0, 0, 0, 1},
{0, 0, 0, 1, 0}};
/*int graph[NODE][NODE] = {{0, 1, 1, 1, 1},
{1, 0, 1, 0, 0},
{1, 1, 0, 0, 0},
{1, 0, 0, 0, 1},
{1, 0, 0, 1, 0}};*/ //uncomment to check Euler Circuit as well as path
/*int graph[NODE][NODE] = {{0, 1, 1, 1, 0},
{1, 0, 1, 1, 0},
{1, 1, 0, 0, 0},
{1, 1, 0, 0, 1},
{0, 0, 0, 1, 0}};*/ //Uncomment to check Non Eulerian Graph
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;
}
int isEulerian() {
if(isConnected() == false) //when graph is not connected
return 0;
vector<int> degree(NODE, 0);
int oddDegree = 0;
for(int i = 0; i<NODE; i++) {
for(int j = 0; j<NODE; j++) {
if(graph[i][j])
degree[i]++; //increase degree, when connected edge found
}
if(degree[i] % 2 != 0) //when degree of vertices are odd
oddDegree++; //count odd degree vertices
}
if(oddDegree > 2) //when vertices with odd degree greater than 2
return 0;
return 1; //when oddDegree is 0, it is Euler circuit, and when 2, it is Euler path
}
int main() {
if(isEulerian() != 0) {
cout << "The graph has Eulerian path." << endl;
} else {
cout << "The graph has No Eulerian path." << endl;
}
}輸出
The graph has Eulerian path.
更新於：2019年7月30日
994 次瀏覽
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