HDU 3018 Ant Trip (Euler path)

Title Address: HDU 3018

For each point, the number of degrees is odd for each connected branch, and then the number of times required is number/2. Note that the isolated points cannot be counted.

The code is as follows:

#include <iostream> #include <string.h> #include <math.h> #include <queue> #include < algorithm> #include <stdlib.h> #include <map> #include <set> #include <stdio.h>using namespace std; #define LL Long long#define pi ACOs ( -1.0) const int Mod=1e9+7;const int Inf=0x3f3f3f3f;const double eqs=1e-9; const int Maxn=100000+10;int HEAD[MAXN], CNT, VIS[MAXN], DEG[MAXN], sum, ans;struct node{int i, V, next;}        edge[4*maxn];void Add (int u, int v) {edge[cnt].v=v;        Edge[cnt].next=head[u]; head[u]=cnt++;}        void Dfs (int u) {if (deg[u]&1) sum++;                for (int i=head[u];i!=-1;i=edge[i].next) {int v=edge[i].v;                        if (!vis[v]) {vis[v]=1;                DFS (v);        }}}void init () {memset (head,-1,sizeof (head));        cnt=0;        memset (vis,0,sizeof (VIS));        memset (deg,0,sizeof (deg)); Ans=0;}     int main () {int n, m, I, J, u, V;   while (scanf ("%d%d", &n,&m)!=eof) {init ();                        while (m--) {scanf ("%d%d", &u,&v);                        if (u==v) continue;                        Add (U,V);                        Add (V,u);                deg[u]++;d eg[v]++; } for (i=1;i<=n;i++) {if (!vis[i]&&[i]) {sum                                = 0;                                DFS (i);                                if (sum) ANS+=SUM/2;                        else ans+=1;        }} printf ("%d\n", ans); } return 0;}

HDU 3018 Ant Trip (Euler path)