關於臨接表2種方式實現的測驗（Adjacency List）

#include<cstdio>#include<iostream>#include<cstring>const int maxn = 40;int u[maxn],v[maxn],next[maxn],first[maxn];int n,m;void read_graph(){    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++)        first[i]=-1;    for(int e=0;e<m;e++)    {        scanf("%d%d",&u[e],&v[e]);        next[e]=first[u[e]];        first[u[e]]=e;    }}int main(){    read_graph();    for(int i=1;i<=n;i++)    {        printf("Start from %d:\n",i);        for(int id=first[i];id!=-1;id=next[id])        {            printf("Edge:%d st->:%d et->:%d\n",id,u[id],v[id]);        }        printf("\n");    }    return 0;}

#include<cstring>#include<iostream>#include<cstdio>#include<algorithm>using namespace std;const int maxn = 100;int n;struct edge{    int u,v,w,next;};edge E[maxn];//determined by the scale of edgeint LE[maxn];//determined by the scale of pointvoid init(){    for(int i=0;i<maxn;i++)        LE[i]=-1;}void read_graph(){    int u,v,m;    scanf("%d%d",&n,&m);    for(int i=0;i<m;i++){        scanf("%d%d",&u,&v);        E[i].v=v;        E[i].next=LE[u];        LE[u]=i;    }}void print(){    for(int i=1;i<=n;i++){        printf("start from %d:\n",i);        for(int id=LE[i];id!=-1;id=E[id].next){            printf("Edge %d: st:%d et:%d\n",id,i,E[id].v);        }        printf("\n");    }}int main(){    init();    read_graph();    print();    return 0;}