题目链接
题意：问有哪几个匹配一定不会出现在任意一个最大匹配中。
那么，我们不妨直接暴力去枚举每一个匹配，因为带花树的复杂度是的，所以就算加上枚举的复杂度
也是在允许范围内的。
只是注意判断条件，先强制让每次需要的匹配边自动匹配上，然后他们就不允许再换了。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 45, maxM = 260;
int N, M, head[maxN], cnt, forbidden[2];
struct Eddge
{
int nex, to;
Eddge(int a=-1, int b=0):nex(a), to(b) {}
} edge[maxM];
inline void addEddge(int u, int v)
{
edge[cnt] = Eddge(head[u], v);
head[u] = cnt++;
}
inline void _add(int u, int v) { addEddge(u, v); addEddge(v, u); }
bool CanNot(int u, int v) { return u == forbidden[0] || u == forbidden[1] || v == forbidden[0] || v == forbidden[1]; }
struct DHS
{
int match[maxN], pre[maxN], vis[maxN], fa[maxN], tim[maxN], idx, ans;
queue<int> Q;
inline int fid(int x) { return x == fa[x] ? x : fa[x] = fid(fa[x]); }
inline int lca(int x, int y)
{
for(++idx; ; swap(x, y))
{
if(x)
{
x = fid(x);
if(tim[x] == idx) return x;
else
{
tim[x] = idx;
x = pre[match[x]];
}
}
}
}
void blossom(int x, int y, int p)
{
while (fid(x) != p)
{
pre[x] = y; y = match[x];
if (vis[y] == 2) { vis[y] = 1; Q.push(y); }
if (fid(x) == x) fa[x] = p;
if (fid(y) == y) fa[y] = p;
x = pre[y];
}
}
int Aug(int S)
{
for(int i=1; i<=N; i++)
{
vis[i] = pre[i] = 0;
fa[i] = i;
}
while (!Q.empty()) Q.pop();
Q.push(S);vis[S]=1;
while (!Q.empty())
{
int u = Q.front(); Q.pop();
for (int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(CanNot(u, v)) continue;
if (fid(u) == fid(v) || vis[v] == 2) continue;
if (!vis[v])
{
vis[v] = 2; pre[v] = u;
if (!match[v])
{
for (int x = v, lst; x; x = lst)
{
lst = match[pre[x]];
match[x] = pre[x];
match[pre[x]] = x;
}
return 1;
}
vis[match[v]] = 1;
Q.push(match[v]);
}
else
{
int gg = lca(u,v);
blossom(u, v, gg);
blossom(v, u, gg);
}
}
}
return 0;
}
inline int solve()
{
for(int i=1; i<=N; i++)
{
if(!match[i]) ans += Aug(i);
}
return ans;
}
} dhs;
pair<int, int> E[maxM];
vector<int> ans;
inline void init()
{
ans.clear();
dhs.ans = 0; dhs.idx = 0; cnt = 0; forbidden[0] = forbidden[1] = -1;
for(int i=1; i<=N; i++) { head[i] = -1; dhs.match[i] = dhs.tim[i] = 0; }
}
inline void each_Init()
{
dhs.ans = 0; dhs.idx = 0;
for(int i=1; i<=N; i++) { dhs.match[i] = dhs.tim[i] = 0; }
}
int main()
{
while(scanf("%d%d", &N, &M) != EOF)
{
init();
for(int i=1, u, v; i<=M; i++)
{
scanf("%d%d", &u, &v);
_add(u, v);
E[i] = make_pair(u, v);
}
int max_ans = dhs.solve(), tmp = 0;
for(int i=1; i<=M; i++)
{
each_Init();
forbidden[0] = E[i].first; forbidden[1] = E[i].second;
if(forbidden[0] > forbidden[1]) swap(forbidden[0], forbidden[1]);
tmp = dhs.solve();
if(max_ans > tmp + 1) ans.push_back(i);
}
int len = (int)ans.size();
printf("%d\n", len);
for(int i=0; i<len; i++) printf("%d%c", ans[i], i == len - 1 ? '\n' : ' ');
if(!len) puts("");
}
return 0;
}