Hard Life UVA

题意：

　　rt

解析：

　　我用的第二种方法。。。

　　s向所有的边连权值为1的边

　　所有的点向t连权值为mid的边

　　如果存在u -  > v  则边向u和v分别连一条权值为INF的边

　　二分mid

　　用dfs从s 顺着边走标记点

　　然后输出1 - n种被标记的点即可

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d
", a)
#define plld(a) printf("%lld
", a)
#define pc(a) printf("%c
", a)
#define ps(a) printf("%s
", a)
#define MOD 2018
#define eps 1e-7
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 10010, INF = 0x7fffffff;
int n, m, s, t;
vector<int> f, g;
struct edge
{
    int u, v;
}Edge[maxn];
int head[maxn], cur[maxn], vis[maxn], d[maxn], cnt, nex[maxn << 1];
int ans;
struct node
{
    int u, v;
    double c;
}Node[maxn << 1];

void add_(int u, int v, double c)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].c = c;
    nex[cnt] = head[u];
    head[u] = cnt++;
}

void add(int u, int v, double c)
{
    add_(u, v, c);
    add_(v, u, 0);
}

bool bfs()
{
    queue<int> Q;
    mem(d, 0);
    Q.push(s);
    d[s] = 1;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i = head[u]; i != -1; i = nex[i])
        {
            int v = Node[i].v;
            if(!d[v] && Node[i].c > 0)
            {
                d[v] = d[u] + 1;
                Q.push(v);
                if(v == t) return 1;
            }
        }
    }
    return d[t] != 0;
}

double dfs(int u, double cap)
{
    double ret = 0;
    if(u == t || abs(cap) < eps)
        return cap;
    for(int &i = cur[u]; i != -1; i = nex[i])
    {
        int v = Node[i].v;
        if(d[v] == d[u] + 1 && Node[i].c > 0)
        {
            double V = dfs(v, min(cap, Node[i].c));
            Node[i].c -= V;
            Node[i ^ 1].c += V;
            ret += V;
            cap -= V;
            if(cap == 0) break;
        }
    }
    if(cap > 0) d[u] = -1;
    return ret;
}

double Dinic()
{
    double ans = 0;
    while(bfs())
    {
        memcpy(cur, head, sizeof head);
        ans += dfs(s, INF);
    }
    return ans;
}

void build(double mid)
{
    mem(head, -1), cnt = 0;
    rap(i, 1, m)
    {
        add(s, i, 1);
        add(i, m + Edge[i].u, INF);
        add(i, m + Edge[i].v, INF);
    }
    rap(i, 1, n)
    {
        f.push_back(cnt);
    //    cout << mid << endl;
        add(m + i, t, mid);
    }
}

void f_dfs(int u)
{
    for(int i = head[u]; i != -1; i = nex[i])
    {
        int v = Node[i].v;
        if(!vis[v] && Node[i].c > eps)
        {
            vis[v] = 1, f_dfs(v);
            if(v - m >= 1 && v - m <= n) ans++;
        }
    }    
}



int main()
{
    while(scanf("%d%d", &n, &m) != EOF)
    {
        ans = 0;
        f.clear();
        g.clear();
        if(m == 0)
        {
            cout << 1 << endl;
            cout << 1 << endl;
            continue;
        }
        mem(head, -1);
        cnt = 0;
        int u, v, sum = m;
        f.clear();
        s = 0, t = n + m + 1;
        rap(i, 1, m)
        {
            rd(Edge[i].u), rd(Edge[i].v);
        }
        double l = 1 / (double) n, r = m;
        while(r - l > (1.0 / n / n))
        {
           // mem(head, -1), cnt = 0;
           // f.clear();
            double mid = (r + l) / (double) 2;
            build(mid);
            if(sum - Dinic() > eps) l = mid;
            else r = mid;
        }
        f.clear();
           build(l);
           Dinic();
           mem(vis, 0);
           f_dfs(s);
        cout << ans << endl;
        for(int i = 1; i <= n; i++)
            if(vis[i + m])
                cout << i << endl;
        
    }

    return 0;
}