**1. Test instructions: Give a simple picture without a direction, ask at least a few strokes to finish drawing all the sides.**

**2. Idea:① first Use and check set to find a few connected components;② If there is only one node in the connected component, then it is 0 strokes;③ in a simple undirected graph, if there is no Euler loop, at least use N/2 pen to paint all sides, N is the number of singularities. **

**3AC Code One (93ms):**

** **

#include <cstdio> #include <cstring> #include <set> #include <vector>using namespace Std;int n,m int Father[100005];int deg[100005];//Each node's degree int vis[100005];int odd_cnt[100005];vector<int> vec;int Find (int a) {int r=a; while (Father[a]!=a) {a=father[a]; } father[r]=a; return A;} inline void Union (int a,int b) {a=find (a); B=find (b);//Don't write father[b] Ah ah!!!!!!!!!!!!!!!! 1 if (a!=b) {father[b]=a; }}int Main () {int ans; while (scanf ("%d%d", &n,&m) ==2) {ans=0; Vec.clear (); for (int i=1; i<=n; i++) {father[i]=i; deg[i]=0; vis[i]=0; odd_cnt[i]=0; } for (int i=0; i<m; i++) {int A, B; scanf ("%d%d", &a,&b); deg[a]++; deg[b]++; Union (A, b); } for (int i=1; i<=n; i++) {father[i]=find (i); if (vis[father[i]]==0) {Vec.push_back (father[i]); Vis[father[i]]=1; } if (deg[i]%2) odd_cnt[father[i]]++; } for (int i=0;i<vec.size (); i++) {int f=vec[i]; if (deg[f]==0) continue; if (odd_cnt[f]==0) ans=ans+1; Else ans=ans+odd_cnt[f]/2;//in a simple undirected connectivity graph, if there is no Euler loop, at least use N/2 pen to paint all sides, N is the number of singularities} printf ("%d\n", ans); } return 0;}

AC Code Two (156MS):

#include <cstdio> #include <cstring> #include <set>using namespace std;struct node{int num; Set<int> St; void Init () {num=0; St.clear (); }};int N,m;int father[100005]; Node son[100005];//the number of Sons of node I son[i].num, then loaded in the set St int deg[100005];//each node's degree int Find (int a) {int r=a; while (Father[a]!=a) {a=father[a]; } father[r]=a; return A;} void Union (int a,int b) {a=find (a); B=find (b); if (a!=b) {father[b]=a; }}int Main () {int ans; while (scanf ("%d%d", &n,&m) ==2) {ans=0; for (int i=1; i<=n; i++) {father[i]=i; son[i]=0; deg[i]=0; Son[i].init (); } for (int i=0; i<m; i++) {int A, B; scanf ("%d%d", &a,&b); deg[a]++; deg[b]++; Union (A, b); } for (int i=1; i<=n; i++) {father[i]=find (i); Son[father[i]].num++; Son[father[i]].st.insert (i); } for (int i=1; i<=n; i++) {if (son[i].num>=2) {set<int>::i Terator it; int cnt=0; For (It=son[i].st.begin (); It!=son[i].st.end (); it++) {if (deg[*it]%2) cnt++; } if (!cnt) ans+=1; Else ans=ans+cnt/2;//in a simple undirected graph, at least the N/2 pen is used to paint all sides, N is the number of singularities} printf ("%d\n", ans); } return 0;}

AC Code Three (78MS):

#include <cstdio> #include <cstring> #include <set>using namespace std;struct node{int son_num; int odd_son_num; void Init () {son_num=0; odd_son_num=0; }};int N,m;int father[100005]; Node son[100005];//node i's singularity son number Son[i]int deg[100005];//each node's degree int Find (int a) {int r=a; while (Father[a]!=a) {a=father[a]; } father[r]=a; return A;} inline void Union (int a,int b) {a=find (a); B=find (b); if (a!=b) {father[b]=a; }}int Main () {int ans; while (scanf ("%d%d", &n,&m) ==2) {ans=0; for (int i=1; i<=n; i++) {father[i]=i; son[i]=0; deg[i]=0; Son[i].init (); } for (int i=0; i<m; i++) {int A, B; scanf ("%d%d", &a,&b); deg[a]++; deg[b]++; Union (A, b); } for (int i=1; i<=n; i++) {father[i]=find (i); son[father[i]].son_num++; if (deg[i]%2) son[father[i]].odd_son_num++; } for (int i=1;i<=n;i++) {if (son[i].son_num>=2) {if (!son[i].odd_ Son_num) Ans+=1; Else ans=ans+son[i].odd_son_num/2;//in a simple undirected graph, at least use the N/2 pen to paint all sides, N is the number of singularities}}/*FO R (int i=1; i<=n; i++) {if (son[i].num>=2) {//printf ("%d\n", son[i].num); Set<int>::iterator it; int cnt=0; For (It=son[i].st.begin (); It!=son[i].st.end (); it++) {//printf ("%d\n", *it); if (deg[*it]%2) cnt++; } if (!cnt) ans+=1; Else ans=ans+cnt/2;//in a simple undirected graph, at least the N/2 pen is used to paint all sides, N is the number of singularities}}*/printf ("%d\n", ans ); } return 0;}

**The above code is only a different way to calculate the number of strokes for each connected component**

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

HDU 3018Ant Trip (a stroke problem, using and checking the connected components of undirected graphs)