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上面是，本题需要输出方案数。
题解
参见国家集训队2007论文集7.胡伯涛《最小割模型在信息学竞赛中的应用》
注意M = 0时，输出1,1
还有就是精度的控制，理论精度应该为1/n/n，我把eps设置为1e-7也过了。
论文中说的二分要根据 h(g) 和 0 的关系，本题中判断 h(g) 和 0 的关系不行，需要判断 h(g) 和 eps 的关系，不太明白为什么。
做法①：
转化为补集的思想求解，方案则是顺着有流量的边走，可以到达的点就是方案中的点。
//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int MAXN = 1e5 + 5;
const int INF = 0x3f3f3f3f;
struct Edge
{
int from, to; double cap, flow; //起点,终点,容量,流量
Edge(int u, int v, double c, double f): from(u), to(v), cap(c), flow(f) {}
};
struct Dinic
{
int n, m, s, t; //结点数,边数(包括反向弧),源点s,汇点t
vector<Edge> edges; //边表。edges[e]和edges[e^1]互为反向弧
vector<int> G[MAXN]; //邻接表，G[i][j]表示结点i的第j条边在edges数组中的序号
int d[MAXN]; //从起点到i的距离（层数差）
int cur[MAXN]; //当前弧下标
bool vis[MAXN]; //BFS分层使用
void init(int n)
{
this->n = n;
edges.clear();
for (int i = 0; i <= n; i++) G[i].clear();
}
void AddEdge(int from, int to, double cap)
{
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
bool BFS()//构造分层网络
{
memset(vis, 0, sizeof(vis));
queue<int> Q;
d[s] = 0;
vis[s] = true;
Q.push(s);
while (!Q.empty())
{
int x = Q.front(); Q.pop();
for (int i = 0; i < G[x].size(); i++)
{
Edge& e = edges[G[x][i]];
if (!vis[e.to] && e.cap > e.flow)
{
vis[e.to] = true;
d[e.to] = d[x] + 1;
Q.push(e.to);
}
}
}
return vis[t];
}
double DFS(int x, double a)//沿阻塞流增广
{
if (x == t || a == 0) return a;
double flow = 0, f;
for (int& i = cur[x]; i < G[x].size(); i++)//从上次考虑的弧
{
Edge& e = edges[G[x][i]];
if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0)//多路增广
{
e.flow += f;
edges[G[x][i]^1].flow -= f;
flow += f;
a -= f;
if (a == 0) break;
}
}
return flow;
}
double MaxFlow(int s, int t)
{
this->s = s; this->t = t;
double flow = 0;
while (BFS())
{
memset(cur, 0, sizeof(cur));
flow += DFS(s, INF);
}
return flow;
}
}solve;
const int maxn = 1005;
const double eps = 1e-7;
int n, m, u[maxn], v[maxn], d[maxn], ans;
bool vis[maxn];
bool check(double x)
{
int s = 0, t = n+1;
solve.init(t);
double U = m;
for (int i = 1; i <= n; i++)
{
solve.AddEdge(s, i, U);
solve.AddEdge(i, t, U+2*x-d[i]);
}
for (int i = 1; i <= m; i++)
{
solve.AddEdge(u[i], v[i], 1);
solve.AddEdge(v[i], u[i], 1);
}
double MF = solve.MaxFlow(s, t);
return (U*n-MF)/2 >= eps;
}
void dfs(int u)
{
ans++;
vis[u] = true;
for (int i = 0; i < solve.G[u].size(); i++)
{
Edge& e = solve.edges[solve.G[u][i]];
if (e.cap > e.flow && !vis[e.to]) dfs(e.to);//顺着残留容量不为0的点走
}
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= m; i++)
{
scanf("%d%d", &u[i], &v[i]);
d[u[i]]++; d[v[i]]++;
}
if (m == 0) { printf("1\n1\n"); return 0; }
double l = 1.0/n, r = m;
while (r-l > eps)
{
double mid = (l+r)/2;
if (check(mid)) l = mid;
else r = mid;
}
check(l);
memset(vis, 0, sizeof(vis));
dfs(0);
printf("%d\n", ans-1);
for (int i = 1; i <= n; i++)
if (vis[i]) printf("%d\n", i);
return 0;
}
/*
5 6
1 5
5 4
4 2
2 5
1 2
3 1
*/
做法②：
转化为最大权闭合子图，最后一次层次图中的的点即为被选择的点或边，由于最后需要用层次图判断，所以记得清空层次图。
//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int MAXN = 1e5 + 5;
const int INF = 0x3f3f3f3f;
struct Edge
{
int from, to; double cap, flow; //起点,终点,容量,流量
Edge(int u, int v, double c, double f): from(u), to(v), cap(c), flow(f) {}
};
struct Dinic
{
int n, m, s, t; //结点数,边数(包括反向弧),源点s,汇点t
vector<Edge> edges; //边表。edges[e]和edges[e^1]互为反向弧
vector<int> G[MAXN]; //邻接表，G[i][j]表示结点i的第j条边在edges数组中的序号
int d[MAXN]; //从起点到i的距离（层数差）
int cur[MAXN]; //当前弧下标
bool vis[MAXN]; //BFS分层使用
void init(int n)
{
this->n = n;
edges.clear();
for (int i = 0; i <= n; i++) G[i].clear();
}
void AddEdge(int from, int to, double cap)
{
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
bool BFS()//构造分层网络
{
memset(vis, 0, sizeof(vis));
memset(d, 0, sizeof(d));//因为输出方案要用，所以清空层次图
queue<int> Q;
d[s] = 0;
vis[s] = true;
Q.push(s);
while (!Q.empty())
{
int x = Q.front(); Q.pop();
for (int i = 0; i < G[x].size(); i++)
{
Edge& e = edges[G[x][i]];
if (!vis[e.to] && e.cap > e.flow)
{
vis[e.to] = true;
d[e.to] = d[x] + 1;
Q.push(e.to);
}
}
}
return vis[t];
}
double DFS(int x, double a)//沿阻塞流增广
{
if (x == t || a == 0) return a;
double flow = 0, f;
for (int& i = cur[x]; i < G[x].size(); i++)//从上次考虑的弧
{
Edge& e = edges[G[x][i]];
if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0)//多路增广
{
e.flow += f;
edges[G[x][i]^1].flow -= f;
flow += f;
a -= f;
if (a == 0) break;
}
}
return flow;
}
double MaxFlow(int s, int t)
{
this->s = s; this->t = t;
double flow = 0;
while (BFS())
{
memset(cur, 0, sizeof(cur));
flow += DFS(s, INF);
}
return flow;
}
}solve;
const int maxn = 1005;
const double eps = 1e-7;
int n, m, u[maxn], v[maxn], d[maxn];
bool check(double x)
{
double tot = 0;
int s = 0, t = n+m+1;
solve.init(t);
for (int i = 1; i <= n; i++) solve.AddEdge(i, t, x);
for (int i = 1; i <= m; i++)
{
solve.AddEdge(s, i+n, 1); tot += 1;
solve.AddEdge(i+n, u[i], INF);
solve.AddEdge(i+n, v[i], INF);
}
double MF = solve.MaxFlow(s, t);
return tot-MF >= eps;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= m; i++)
{
scanf("%d%d", &u[i], &v[i]);
d[u[i]]++; d[v[i]]++;
}
if (m == 0) { printf("1\n1\n"); return 0; }
double l = 1.0/n, r = m;
while (r-l > eps)
{
double mid = (l+r)/2;
if (check(mid)) l = mid;
else r = mid;
}
check(l);
int ans = 0;
for (int i = 1; i <= n; i++)
if (solve.d[i]) ans++;
printf("%d\n", ans);
for (int i = 1; i <= n; i++)
if (solve.d[i]) printf("%d ", i);
return 0;
}
/*
5 6
1 5
5 4
4 2
2 5
1 2
3 1
*/