假设图用邻接表表示，设计一个算法，输出从顶点vi到顶点vj的所有简单路径

#include <iostream>]
#include <string.h>
#define maxsize 10
typedef struct node{
    int data;
    struct node *next;
}node ,*pnode;


pnode buynode(int x)
{
    pnode tmp=(pnode) malloc(sizeof (node));
    tmp->data=x;
    tmp->next= nullptr;
    return tmp;
}

pnode * graph(int point)//结点数量
{
    pnode * g=(pnode*) malloc(sizeof (pnode)*point);
    for(int i=0;i<point;i++)
        g[i]= buynode(i);
    return g;
}
void insert(pnode* g,int p1,int p2)
{
    pnode tmp=g[p1];
    while(tmp->next&&tmp->next->data<p2) tmp=tmp->next;
    pnode ne=tmp->next;
    tmp->next= buynode(p2);
    tmp->next->next=ne;

    tmp=g[p2];
    while(tmp->next&&tmp->next->data<p1) tmp=tmp->next;
    ne=tmp->next;
    tmp->next= buynode(p1);
    tmp->next->next=ne;
}
void figure1(pnode* g)
{
    insert(g,0,1);
    insert(g,1,2);
    insert(g,2,3);
    insert(g,3,4);
    insert(g,4,0);
    insert(g,0,2);
    insert(g,0,3);

}
void print(pnode *g,int point)
{
    for(int i=0;i<5;i++)
    {
        pnode tmp=g[i];
        while(tmp) {
            printf("%3d",tmp->data);
            tmp=tmp->next;
        }
        puts("");
    }
}

void _dfs(pnode *g,int p1,int p2,int*visited,int*path,int &pointer)
{
    visited[p1]=1;
    path[++pointer]=p1;
    if(p1==p2){
        for(int i=0;i<=pointer;i++) printf("%2d->",path[i]);
        puts("");
        return;
    }
    pnode tmp=g[p1];
    while(tmp->next){
        if(!visited[tmp->next->data])//没有访问过
        {
            _dfs(g,tmp->next->data,p2,visited,path,pointer);
            visited[path[pointer]]=0;
            pointer--;
        }
        tmp=tmp->next;
    }
}
void dfs(pnode* g,int p1,int p2)
{
    int *visited=(int*) malloc(sizeof (int)*maxsize);
    int *path=(int*) malloc(sizeof (int )*maxsize);
    int pointer=-1;
    for(int i=0;i<maxsize;i++) visited[i]=0,path[i]=0;
    printf("the paths are:\n");
    _dfs(g,p1,p2,visited,path,pointer);
}
int main() {
    pnode* g1=graph(5);
    figure1(g1);
    print(g1,5);
    dfs(g1,0,3);
    return 0;
}

测试用的树如下

测试从0到3的全部路径