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("******
")
#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 + 7;
const int maxn = 2e5 + 10;
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] = 1;
    if ( du[u] & 1 ) cnt.push_back ( u );
    for ( int i = 0 ; i < g[u].size() ; i++ ) {
        if ( vis[g[u][i].v] ) continue;
        dfs ( g[u][i].v );
    }
}
void dfs1 ( int u ) {
    for ( int i = 0 ; i < g[u].size() ; i++ ) {
        if ( vis1[abs ( g[u][i].id )] ) continue;
        vis1[abs ( g[u][i].id )] = 1;
        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 = 0;
        mem ( vis, 0 ), mem ( vis1, 0 );
        for ( int i = 1 ; i <= n ; i++ ) path[i].clear(), g[i].clear(), du[i] = 0;
        for ( int i = 1, 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 = 1 ; i <= n ; i++ ) {
            if ( !vis[i] && du[i] ) {
                cnt.clear();
                dfs ( i );
                ans++;
                if ( cnt.size() == 0 ) cnt.push_back ( i );
                for ( int j = 2 ; j < cnt.size() ; j += 2 ) {
                    g[cnt[j]].push_back ( {cnt[j + 1], ++num} );
                    g[cnt[j + 1]].push_back ( {cnt[j], -num} );
                }
                dfs1 ( cnt[0] );
            }
        }
        printf ( "%d
", ans );
        for ( int i = 1 ; i <= ans; i++ ) {
            printf ( "%d", path[i].size() );
            for ( int j = 0 ; j < path[i].size() ; j++ ) printf ( " %d", path[i][j] );
            printf ( "
" );
        }
    }
    return 0 ;
}