Ninth HDU 3926 hand in multiple schools (a special picture homogeneous problem)

The condition that the degree of vertices is less than or equal to 2 is given, that is, the graph can only be a graph composed of rings and links. The vertices are included in the chain, so that you only need to repeat each connected component, record whether they are loops, links, and the number of points, and then sort and compare them.

#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;}