hdu 6311 欧拉回路

题意：求一个图(不一定联通)最小额外连接几条边，使得可以一笔画出来

大致做法

1.找出联通块

2.统计每一个连通块里面度数为奇数的点的个数，

有一个性质 一个图能够用一笔画出来，奇数点的个数不超过2个

if  奇数点的个数==0  或者 ==1 直接找欧拉回路

else 将除去前面两个奇数点外的奇数点依次相连 然后找欧拉回路

然后记录路径

 #include <cstdio>
 #include <cstring>
 #include <queue>
 #include <cmath>
 #include <algorithm>
 #include <set>
 #include <iostream>
 #include <map>
 #include <stack>
 #include <string>
 #include <vector>
 #define  pi acos(-1.0)
 #define  eps 1e-9
 #define  fi first
 #define  se second
 #define  rtl   rt<<1
 #define  rtr   rt<<1|1
 #define  bug         printf("******\n")
 #define  mem(a,b)    memset(a,b,sizeof(a))
 #define  name2str(x) #x
 #define  fuck(x)     cout<<#x" = "<<x<<endl
 #define  f(a)        a*a
 #define  sf(n)       scanf("%d", &n)
 #define  sff(a,b)    scanf("%d %d", &a, &b)
 #define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
 #define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  pf          printf
 #define  FRE(i,a,b)  for(i = a; i <= b; i++)
 #define  FREE(i,a,b) for(i = a; i >= b; i--)
 #define  FRL(i,a,b)  for(i = a; i < b; i++)+
 #define  FRLL(i,a,b) for(i = a; i > b; i--)
 #define  FIN         freopen("data.txt","r",stdin)
 #define  gcd(a,b)    __gcd(a,b)
 #define  lowbit(x)   x&-x
 using namespace std;
 typedef long long  LL;
 typedef unsigned long long ULL;
 const int mod = 1e9 + ;
 const int maxn = 2e5 + ;
 const int INF = 0x3f3f3f3f;
 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 int n, m, ans, du[maxn], vis[maxn], vis1[maxn];
 struct Edge {
     int v, id;
 };
 vector<Edge>g[maxn];
 vector<int>cnt, path[maxn];
 void dfs ( int u ) {
     vis[u] = ;
     if ( du[u] &  ) cnt.push_back ( u );
     for ( int i =  ; i < g[u].size() ; i++ ) {
         if ( vis[g[u][i].v] ) continue;
         dfs ( g[u][i].v );
     }
 }
 void dfs1 ( int u ) {
     for ( int i =  ; i < g[u].size() ; i++ ) {
         if ( vis1[abs ( g[u][i].id )] ) continue;
         vis1[abs ( g[u][i].id )] = ;
         dfs1 ( g[u][i].v );
         if ( abs ( g[u][i].id ) > m ) ans++;
         else path[ans].push_back ( -g[u][i].id );
         //欧拉回路的路径是反的  所以要-号
     }
 }
 int main() {
     while ( ~sff ( n, m ) ) {
         ans = ;
         mem ( vis,  ), mem ( vis1,  );
         for ( int i =  ; i <= n ; i++ ) path[i].clear(), g[i].clear(), du[i] = ;
         for ( int i = , u, v ; i <= m ; i++ ) {
             sff ( u, v );
             g[u].push_back ( {v, i} );
             g[v].push_back ( {u, -i} );
             du[u]++, du[v]++ ;
         }
         int num = m;
         for ( int i =  ; i <= n ; i++ ) {
             if ( !vis[i] && du[i] ) {
                 cnt.clear();
                 dfs ( i );
                 ans++;
                 if ( cnt.size() ==  ) cnt.push_back ( i );
                 for ( int j =  ; j < cnt.size() ; j +=  ) {
                     g[cnt[j]].push_back ( {cnt[j + ], ++num} );
                     g[cnt[j + ]].push_back ( {cnt[j], -num} );
                 }
                 dfs1 ( cnt[] );
             }
         }
         printf ( "%d\n", ans );
         for ( int i =  ; i <= ans; i++ ) {
             printf ( "%d", path[i].size() );
             for ( int j =  ; j < path[i].size() ; j++ ) printf ( " %d", path[i][j] );
             printf ( "\n" );
         }
     }
     return  ;
 }