多校第九場HDU 3926 hand in hand (一種特殊圖的同構問題)

題裡給出了沒點的度都小於等於2這個條件，既圖只能是環和鏈組成的圖，其中點包括在鏈裡，這樣只需遍曆下各個連通分量，記錄他們是否是環還是鏈，以及點的個數，之後在排序對比下即可。

#include <cstdio>#include <algorithm>using namespace std;const int inf =0x3fffffff;const int maxn=10005;struct graph{     int chain[maxn],ring[maxn];     int chain_num,ring_num;}g1,g2;struct Edge {    int v,next;}edge[2*maxn];int head[maxn],cnt;bool vis[maxn];void addedge (int u,int v){     edge[cnt].v=v;     edge[cnt].next=head[u];     head[u]=cnt++;     edge[cnt].v=u;     edge[cnt].next=head[v];     head[v]=cnt++;}bool is_same(){     sort (g1.chain,g1.chain+g1.chain_num);     sort (g2.chain,g2.chain+g2.chain_num);     for (int i=0 ; i<g1.chain_num ; ++i)     if(g1.chain[i]!=g2.chain[i])return false;          sort (g1.ring,g1.ring+g1.ring_num);     sort (g2.ring,g2.ring+g2.ring_num);     for (int i=0 ; i<g1.ring_num ; ++i)     if(g1.ring[i]!=g2.ring[i])return false;     return true;}void init (graph &g){     memset (g.chain , 125 , sizeof(g.chain));     memset (g.ring , 125 , sizeof(g.ring));     g.chain_num=g.ring_num=0;     memset (head , -1 , sizeof(head));     memset (vis , 0 , sizeof(vis));     cnt=0;}bool is_ring;void dfs (int u,int &num,int fath){     vis[u]=true; num++;     for (int p=head[u]; ~p ; p=edge[p].next)     {         int v=edge[p].v;         if(v==fath)continue;         if(vis[v])is_ring=true;         if (!vis[v])         {              dfs(v,num,u);         }     }}int m,n,u,v,i,j;void solve(graph &g){     init(g);     scanf("%d%d",&n,&m);     for (i=0 ; i<m ; ++i)     {         scanf("%d%d",&u,&v);         u--,v--;         addedge(u,v);     }     int count;     for (i=0 ; i<n ; ++i)     {         if (!vis[i])         {               count=1;               is_ring=false;               dfs(i,count,inf);               if(is_ring)g.ring[g.ring_num++]=count;               else g.chain[g.chain_num++]=count;         }     }}int main (){    int cas;    //freopen ("G.txt","r",stdin);    //freopen ("out.txt","w",stdout);    scanf("%d",&cas);    for (int I=1 ; I<=cas ; ++I)    {        solve(g1);        solve(g2);        printf("Case #%d: %s\n",I,is_same()?"YES":"NO");    }    return 0;}