题意：给出一张有向图，每次你可以从图中的任意一点出发，经过若干条边后停止，然后问你最少走几次可以将图中的每条边都走过至少一次，并且要输出方案，这个转化为网络流的话，就相当于 求一个最小流，并且存在下界，即每条边至少走一次。
析：转载：http://blog.csdn.net/sdj222555/article/details/40380423
这上面说的很清楚了，就是先求一遍是正向的，然后再求逆向的，这个逆向就是为了求可以减少的边。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-3;
const int maxn = 100 + 40;
const int maxm = (100000 << 8) + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
int from, to, cap, flow;
};
struct Dinic{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
int d[maxn];
bool vis[maxn];
int cur[maxn];
void init(int n){
this-> n = n;
FOR(i, 0, n) G[i].cl;
edges.cl;
}
void addEdge(int from, int to, int cap){
edges.pb((Edge){from, to, cap, 0});
edges.pb((Edge){to, from, 0, 0});
m = edges.sz;
G[from].pb(m - 2);
G[to].pb(m - 1);
}
bool bfs(){
ms(vis, 0); vis[s] = 1;
d[s] = 0;
queue<int> q; q.push(s);
while(!q.empty()){
int u = q.front(); q.pop();
for(int i = 0; i < G[u].sz; ++i){
Edge &e = edges[G[u][i]];
if(!vis[e.to] && e.cap > e.flow){
d[e.to] = d[u] + 1;
vis[e.to] = 1;
q.push(e.to);
}
}
}
return vis[t];
}
int dfs(int u, int a){
if(u == t || a == 0) return a;
int flow = 0, f;
for(int &i = cur[u]; i < G[u].sz; ++i){
Edge &e = edges[G[u][i]];
if(d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0){
e.flow += f;
edges[G[u][i]^1].flow -= f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
}
int maxflow(int s, int t){
this-> s = s;
this-> t = t;
int flow = 0;
while(bfs()){ ms(cur, 0); flow += dfs(s, INF); }
return flow;
}
};
Dinic dinic;
int in[maxn];
vector<P> edges[maxn];
int main(){
while(scanf("%d", &n) == 1){
ms(in, 0);
int s = 0, t = n + 1;
dinic.init(t + 4);
int ans = 0;
for(int i = 1; i <= n; ++i){
int x; scanf("%d", &x);
while(x--){
int y; scanf("%d", &y);
++in[y]; --in[i];
dinic.addEdge(i, y, INF);
}
edges[i].cl;
}
int sum = 0;
for(int i = 1; i <= n; ++i)
if(in[i] > 0) dinic.addEdge(s, i, in[i]), ans += in[i];
else dinic.addEdge(i, t, -in[i]);
printf("%d\n", ans -= dinic.maxflow(s, t));
for(int i = 1; i <= n; ++i){
for(int j = 0; j < dinic.G[i].sz; ++j){
Edge &e = dinic.edges[dinic.G[i][j]];
if(e.from != i || e.to == t || e.cap == 0) continue;
edges[i].pb(P(e.to, e.flow + 1));
in[e.to] += e.flow;
}
}
for(int i = 1; i <= n; ++i) while(in[i] < 0){
++in[i];
vector<int> ans; ans.push_back(i);
int u = i;
while(1){
bool ok = true;
for(int j = 0; j < edges[u].sz; ++j){
if(edges[u][j].se == 0) continue;
ok = false;
--edges[u][j].se;
u = edges[u][j].fi;
ans.push_back(u);
break;
}
if(ok) break;
}
for(auto &it : ans) printf("%d%c", it, " \n"[it == ans.back()]);
}
}
return 0;
}
