上一篇文章當中介紹了Fleury演算法是什麼以及演算法流程，本篇文章將介紹如何用程式碼來實現求解歐拉回路。廢話不多說先來三個例子，這三個例子分別是兩個尤拉圖和一個非尤拉圖。

  非尤拉圖：

圖 1  

  尤拉圖一：

圖 2 (此圖來自《離散數學教程》p135)

  尤拉圖二：(這是一個七個點的圖，每一個點都是6條邊)

圖 3

  現在我們來開始編碼，首先，建立一個二維陣列來儲存圖，然後是輸入點的個數和邊的個數，以及哪兩個點有邊。以下是程式碼片段：

#include<stdio.h>

int eulerGraph[1000][1000] = {{0}};//用來儲存圖，全域性的二維陣列

int V,E;//邊以及點的個數
 

printf("please input the number of V (no more than 1000): \n");
scanf("%d",&V);
printf("please input the number of E : \n");
scanf("%d",&E);
printf("please input E : \n");
int i,j;
//Init eulerGraph
	
for(i = 0; i < V; i++) {
	for(j = 0; j < V; j++) {
		eulerGraph[i][j] = 0;
	}
}
int m,n;
//create graph
for (i = 0; i < E; i++) {
	scanf("%d %d",&m,&n);
	if ((m >= V) || (n >= V)) {
		printf("cannot more than V\n");
		return 1;
	}
	eulerGraph[m][n] = eulerGraph[n][m] = 1;
}
 

	for(i = 0; i < V; i++) {
		for(j = 0; j < V; j++) {
			printf("%d ",eulerGraph[i][j]);
		}
		printf("\n");
	}

if (!isEulerGraph()) {
		
		printf("the graph is not euler graph!\n");
		return 1;
	}

int  isEulerGraph() {
	int i,j;
	int count = 0;
	for (i = 0; i < V; i++) {
		for (j = 0; j < V; j++) {
			if (1 == eulerGraph[i][j]) 
				count++;
		}
		if (0 != count % 2 )
			return 0;
		count = 0;
	}
	return 1;
}
 

fleury();

void fleury () {
	//set a start ,v0
	int temp[1000] ={0};
	printf("%d -> ",temp[0]);
	int i = 0;
	int j = 0;
	int top = 0;
	int bridge = 1;
	int tmp;
	for (i = 0; i < E; i++) {//Traversal all E
		for (j = 0; j < V; j++) {
			if(1 == eulerGraph[temp[top]][j]) {
				eulerGraph[temp[top]][j] = eulerGraph[j][temp[top]] = 0;
				//if the E is bridge,take a signal in this E,else delete this E
				if(isBridge(temp[top],j)) {
					tmp = j;		
					eulerGraph[temp[top]][j] = eulerGraph[j][temp[top]] = 1;
				}
				else {
					top++;
					temp[top] = j;
					bridge = 0;
					printf("%d -> ",j);
					break;
				}
					
			}
		}
		//if bridge == 1, no another E in this V that the E is not bridge
		if (bridge) {
			eulerGraph[temp[top]][tmp] = eulerGraph[tmp][temp[top]] = 0;
			top++;
			temp[top] = tmp;
			printf("%d -> ",tmp);
		}
		bridge = 1;
	}
//	for (i = 0; i < (E - 1); i++) {
//		printf("%d -> ",temp[i]);
//	}
//	printf ("%d",temp[E-1]);
}

首先，建立一個數組用來儲存要判斷的Vi能夠到達的點，只要數組裡面的點未遍歷完那麼就把這些點能夠到達的點，並且在陣列中還未遍歷過加進來。以下是原始碼：

int isBridge (int m, int k) {
	int temp[1000] = {-1};
	int i = 0;
	for (i = 0; i < 1000; i++) {
		temp[i] = -1;
	}
	int tmp = 0;
	temp[tmp] = m;
	int n = 0;
	//判斷從m->k是否還有其他路徑
	for (tmp = 0; temp[tmp] != -1; tmp++ ) {		
		for (i = 0; i < V; i++) {
			
			if (eulerGraph[temp[tmp]][i] == 1 && i == k) {
				return 0;//because is exist another way to k,so m->k is not bridge  
			}
			if (eulerGraph[temp[tmp]][i] == 1 && isExist(temp,i) == 0) {//將當前的點有邊的點，前面沒有到過的點加入到陣列當中
				n++;
				temp[n] = i;//take new V into temp array
			}
		}
	}
	//not found k,so m->k is bridge
	return 1;
}

int isExist (int *temp,int m) {
	int i = 0;
	while (temp[i] != -1) {
		if (temp[i] == m)
			return 1;
		i++;
	}
	return 0;
}

測試非尤拉圖：

測試尤拉圖一：

測試尤拉圖二：(本人有點懶，所以就不打算使用手動輸入的方式嘍，採用輸入重定向的方式來進行輸入，大家平時測試的時候也可以這樣做，畢竟比較方便一點)

輸入：

7
21
0 1
0 2
0 3
0 4
0 5
0 6
1 2
1 3
1 4
1 5
1 6
2 3
2 4
2 5
2 6
3 4
3 5
3 6
4 5
4 6
5 6

輸出：

以上就是我對於Fleury演算法的理解，如果覺得還可以，請點個贊或者在下方評論，以下是我的原始碼：

/*************************************************************************
    > File Name: Fleury.c
    > Author: pengjinghui
    > Mail: [email protected] 
    > Created Time: 2015年09月24日 星期四 21時56分27秒
 ************************************************************************/

#include<stdio.h>

int eulerGraph[1000][1000] = {{0}};//用來儲存圖，全域性的二維陣列

int V,E;//邊以及點的個數

int  isEulerGraph() {
	int i,j;
	int count = 0;
	for (i = 0; i < V; i++) {
		for (j = 0; j < V; j++) {
			if (1 == eulerGraph[i][j]) 
				count++;
		}
		if (0 != count % 2 )
			return 0;
		count = 0;
	}
	return 1;
}
int isExist (int *temp,int m) {
	int i = 0;
	while (temp[i] != -1) {
		if (temp[i] == m)
			return 1;
		i++;
	}
	return 0;
}
int isBridge (int m, int k) {
	int temp[1000] = {-1};
	int i = 0;
	for (i = 0; i < 1000; i++) {
		temp[i] = -1;
	}
	int tmp = 0;
	temp[tmp] = m;
	int n = 0;
	//判斷從m->k是否還有其他路徑
	for (tmp = 0; temp[tmp] != -1; tmp++ ) {		
		for (i = 0; i < V; i++) {
			
			if (eulerGraph[temp[tmp]][i] == 1 && i == k) {
				return 0;//because is exist another way to k,so m->k is not bridge  
			}
			if (eulerGraph[temp[tmp]][i] == 1 && isExist(temp,i) == 0) {//將當前的點有邊的點，前面沒有到過的點加入到陣列當中
				n++;
				temp[n] = i;//take new V into temp array
			}
		}
	}
	//not found k,so m->k is bridge
	return 1;
}
void fleury () {
	//set a start ,v0
	int temp[1000] ={0};
	printf("%d -> ",temp[0]);
	int i = 0;
	int j = 0;
	int top = 0;
	int bridge = 1;
	int tmp;
	for (i = 0; i < E; i++) {//Traversal all E
		for (j = 0; j < V; j++) {
			if(1 == eulerGraph[temp[top]][j]) {
				eulerGraph[temp[top]][j] = eulerGraph[j][temp[top]] = 0;
				//if the E is bridge,take a signal in this E,else delete this E
				if(isBridge(temp[top],j)) {
					tmp = j;		
					eulerGraph[temp[top]][j] = eulerGraph[j][temp[top]] = 1;
				}
				else {
					top++;
					temp[top] = j;
					bridge = 0;
					printf("%d -> ",j);
					break;
				}
					
			}
		}
		//if bridge == 1, no another E in this V that the E is not bridge
		if (bridge) {
			eulerGraph[temp[top]][tmp] = eulerGraph[tmp][temp[top]] = 0;
			top++;
			temp[top] = tmp;
			printf("%d -> ",tmp);
		}
		bridge = 1;
	}
//	for (i = 0; i < (E - 1); i++) {
//		printf("%d -> ",temp[i]);
//	}
//	printf ("%d",temp[E-1]);
}

int main () {
	printf("please input the number of V (no more than 1000): \n");
	scanf("%d",&V);
	printf("please input the number of E : \n");
	scanf("%d",&E);
	printf("please input E : \n");
	int i,j;
	//Init eulerGraph
	
	for(i = 0; i < V; i++) {
		for(j = 0; j < V; j++) {
			eulerGraph[i][j] = 0;
		}
	}
	int m,n;
	//create graph
	for (i = 0; i < E; i++) {
		scanf("%d %d",&m,&n);
		if ((m >= V) || (n >= V)) {
			printf("cannot more than V\n");
			return 1;
		}
		eulerGraph[m][n] = eulerGraph[n][m] = 1;
	}
	
	for(i = 0; i < V; i++) {
		for(j = 0; j < V; j++) {
			printf("%d ",eulerGraph[i][j]);
		}
		printf("\n");
	}
	if (!isEulerGraph()) {
		
		printf("the graph is not euler graph!\n");
		return 1;
	}
	fleury();
}