Judge loop in directed graph

1 深度優先方法

首先須要更改矩陣初始化函數init_graph()

      而後咱們須要初始化vist標記數組

深度優先訪問圖，而後根據是否存在back edge判斷是否存在環路

算法以下：

#include <iostream>
using namespace std;

#define MAX_VERTEX_NUM 128
enum color{WHITE, GRAY = 1, BLACK};
bool M[MAX_VERTEX_NUM][MAX_VERTEX_NUM];
int colour[MAX_VERTEX_NUM];
int dfsNum[MAX_VERTEX_NUM], num;
int indegree[MAX_VERTEX_NUM];
int vexnum, edgenum;
bool loop;

void init_graph(){
    cout<<"enter vertex number:"<<endl;
    cin>>vexnum;
    cout<<"enter edge number:"<<endl;
    cin>>edgenum;

    int i, j;
    while(edgenum){
        cout<<"add new edge:"<<endl;
        cin>>i>>j;
        M[i - 1][j - 1] = true;
        //initialize in vertex degree
        indegree[i - 1]++;
        indegree[j - 1]++;
        edgenum--;
    }
}

void dfs(int u, int p){
    colour[u] = GRAY;
    dfsNum[u] = num++;
    cout<<"old relation:"<<endl;
    cout<<"child: "<<u + 1<<" parent :"<<p + 1<<endl;
    for( int v = 0; v < vexnum; v++){
        if(M[u][v] && v != p){
            if(colour[v] == WHITE){
                cout<<"new relation:"<<endl;
                cout<<"parent "<<u + 1<<"(color: "<<colour[u]<<")"
                    <<"and child "<<v + 1<<"(color: "<<colour[v]<<")"<<endl;
                dfs(v, u);
                cout<<"parent "<<u + 1<<"(color: "<<colour[u]<<")"
                    <<"and child "<<v + 1<<"(color: "<<colour[v]<<")"<<endl;
            }
            else if(colour[v] == GRAY){
                   cout<<"back edge between"<<u + 1<<" and "<<v + 1<<endl;
                   loop = true;
                  // break;
            }
            else if(colour[v] == BLACK){
                cout<<"cross edge between"<<u + 1<<" and"<<v + 1<<endl;;
                loop = true;
               // break;
            }
        }
    }
    colour[u] = BLACK;
}
void print_dfs_num(){
    for(int v = 0; v < vexnum; v++)
        cout<<dfsNum[v]<<" ";
}


int main()
{
    init_graph();
    dfs(0, -1);
    print_dfs_num();
    cout<<endl;
    if(loop)
        cout<<"There is loop in graph!"<<endl;

    int ch;
    cin>>ch;
    return 0;
}

版權聲明：本文爲博主原創文章，未經博主容許不得轉載。