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

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

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