題意：

給你一個二分圖。

問最少保留哪些邊，使得每個點的度數至少為k。k從0到mindegree。

1 <= n1, n2 <= 2000. m <= 2000

題解：

網路流。

考慮逆向思考。

我們用源點連第一部分的點，匯點連第二部分的點。

對於某一個k，這些邊的容量是點的度數-k，代表不選的原邊。

然後第一部分的點用原邊連第二部分的點，容量為1。

因為求的是最大流，所以不選邊數的最大，選的邊數最小，並且點度數都至少為k，正好符合題目條件。

對於k-1，我們只要把邊的容量+1，然後在殘餘網路上再跑網路流即可。

所以k從mindegree列舉到0。複雜度我算不清楚，題解上說是

程式碼：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <bitset>
#include <map>
#include <vector>
#include <stack>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <ctime>
#ifdef LOCAL
#define debug(x) cout<<#x<<" = "<<(x)<<endl;
#else
#define debug(x) 1;
#endif

#define chmax(x,y) x=max(x,y)
#define chmin(x,y) x=min(x,y)
#define lson id<<1,l,mid
#define rson id<<1|1,mid+1,r
#define lowbit(x) x&-x
#define mp make_pair
#define pb push_back
#define fir first
#define sec second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, int> pii;

const int MOD = 1e9 + 7;
const double PI = acos (-1.);
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 2e5 + 5;

const int inf = 0x3f3f3f3f;
const int MX = 4111;
const int MXE = 1e5;
struct MaxFlow {
    int flow;
    struct Edge {
        int v, w, nxt;
    } edge[MXE];
    int tot, num, s, t;
    int head[MX];
    void init() {
        memset(head, -1, sizeof(head));
        tot = flow = 0;
    }
    void add(int u, int v, int w) {
        edge[tot].v = v;
        edge[tot].w = w;
        edge[tot].nxt = head[u];
        head[u] = tot++;

        edge[tot].v = u;
        edge[tot].w = 0;
        edge[tot].nxt = head[v];
        head[v] = tot++;
    }

    int  d[MX], vis[MX], gap[MX];
    void bfs() {
        memset(d, 0, sizeof(d));
        memset(gap, 0, sizeof(gap));
        memset(vis, 0, sizeof(vis));
        queue<int>q;
        q.push(t);
        vis[t] = 1;
        while (!q.empty()) {
            int u = q.front();
            q.pop();
            for (int i = head[u]; ~i; i = edge[i].nxt) {
                int v = edge[i].v;
                if (!vis[v]) {
                    d[v] = d[u] + 1;
                    gap[d[v]]++;
                    q.push(v);
                    vis[v] = 1;
                }
            }
        }
    }

    int last[MX];
    int dfs(int u, int f) {
        if (u == t) return f;
        int sap = 0;
        for (int i = last[u]; ~i; i = edge[i].nxt) {
            int v = edge[i].v;
            if (edge[i].w > 0 && d[u] == d[v] + 1) {
                last[u] = i;
                int tmp = dfs(v, min(f - sap, edge[i].w));
                edge[i].w -= tmp;
                edge[i ^ 1].w += tmp;
                sap += tmp;
                if (sap == f) return sap;
            }
        }
        if (d[s] >= num) return sap;
        if (!(--gap[d[u]])) d[s] = num;
        ++gap[++d[u]];
        last[u] = head[u];
        return sap;
    }

    int solve(int st, int ed, int n) {
        //flow = 0;
        num = n;
        s = st;
        t = ed;
        bfs();
        memcpy(last, head, sizeof(head));
        while (d[s] < num) flow += dfs(s, inf);
        return flow;
    }
} F;

int eid[MAXN];
int in[MAXN];
vector<int> ans[MAXN];

int main() {
#ifdef LOCAL
    freopen ("input.txt", "r", stdin);
#endif
    int n1, n2, m;
    F.init();
    scanf("%d %d %d", &n1, &n2, &m);
    int st = 0, en = 4050, mindeg = INF;
    for (int i = 1; i <= m; i++) {
        int u, v;
        scanf("%d %d", &u, &v);
        in[u]++;
        in[v + n1]++;
        F.add(u, v + n1, 1);
        eid[i + n1 + n2] = F.tot - 2;
    }

    for (int i = 1; i <= n1; i++) chmin(mindeg, in[i]);
    for (int i = 1; i <= n2; i++) chmin(mindeg, in[i + n1]);
    for (int i = 1; i <= n1; i++) {
        F.add(st, i, in[i] - mindeg);
        eid[i] = F.tot - 2;
    }
    for (int i = 1; i <= n2; i++) {
        F.add(i + n1, en, in[i + n1] - mindeg);
        eid[i + n1] = F.tot - 2;
    }
    puts("0");
    for (int i = mindeg; i > 0; i--) {
        F.solve(st, en, en + 1);
        for (int j = 1; j <= n1; j++) F.edge[eid[j]].w++;
        for (int j = 1; j <= n2; j++) F.edge[eid[j + n1]].w++;
        for (int j = 1; j <= m; j++) if(F.edge[eid[j + n1 + n2]].w) ans[i].pb(j);
    }
    for (int i = 1; i <= mindeg; i++) {
        printf("%d ", ans[i].size());
        for (int j : ans[i]) printf("%d ", j);
        puts("");
    }
    return 0;
}