CodeForces - 976F Minimal k-covering

Description

给你一张左边 \(n_1\) 个点，右边 \(n_2\) 个点， \(m\) 条边的二分图。对于每一个 \(0\le k\le minDeg\) ，求选取哪些边可以使每个点的度数都不小于 \(k\) 。

\(1\le n_1,n_2\le 2000\) ， \(m\le 2000\)

Solution

大力建模谁都会系列，多组询问会炸。

于是建边就建流量为 \(deg[i]-k\) 的边，每次增加流量即可。

#include<bits/stdc++.h>
using namespace std;

template <class T> inline void read(T &x) {
    x = 0; static char ch = getchar(); for (; ch < '0' || ch > '9'; ch = getchar());
    for (; ch >= '0' && ch <= '9'; ch = getchar()) (x *= 10) += ch - '0';
}

#define N 5001
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define INF 0x3f3f3f3f

int S, T, head[N], cur[N], tot = 1, q[N], dep[N];
struct edge { int v, c, next; }e[100001];
inline void insert(int u, int v, int c) { e[++tot].v = v, e[tot].c = c, e[tot].next = head[u]; head[u] = tot; }
inline void add(int u, int v, int c) { insert(u, v, c), insert(v, u, 0); }
inline bool bfs() {
    memset(dep, 0, sizeof dep); dep[S] = 1;
    int l = 1, r = 1; q[1] = S;
    while (l <= r) {
        int u = q[l++];
        for (int i = head[u], v; i; i = e[i].next) if (e[i].c && !dep[v = e[i].v]) {
            dep[v] = dep[u] + 1, q[++r] = v;
            if (!(v ^ T)) return 1;
        }
    }
    return 0;
}
int dfs(int u, int dist) {
    if (u == T) return dist;
    int ret = 0;
    for (int &i = head[u], v; i; i = e[i].next) if (dep[v = e[i].v] == dep[u] + 1 && e[i].c) {
        int d = dfs(v, min(dist - ret, e[i].c));
        e[i].c -= d, e[i ^ 1].c += d, ret += d;
        if (ret == dist) return dist;
    }
    if (!ret) dep[u] = -1;
    return ret;
}
inline void cpy() { rep(i, S, T) cur[i] = head[i]; }
inline void rec() { rep(i, S, T) head[i] = cur[i]; }
int dinic() { int ret = 0; cpy(); while (bfs()) ret += dfs(S, INF), rec(); return ret; }

int nu, nv, n, m, deg[N], minDeg = INF;
vector<int> ans[N];
struct Data { int u, v; }a[N];

int main() {
    read(nu), read(nv), read(m), n = nu + nv, T = n + 1;
    rep(i, 1, m) read(a[i].u), read(a[i].v), a[i].v += nu, deg[a[i].u]++, deg[a[i].v]++;
    rep(i, 1, n) minDeg = min(minDeg, deg[i]);
    rep(i, 1, nu) add(S, i, deg[i] - minDeg - 1);
    rep(i, nu + 1, n) add(i, T, deg[i] - minDeg - 1);
    int tmp = tot;
    rep(i, 1, m) add(a[i].u, a[i].v, 1);
    for (int i = minDeg; i >= 0; i--) {
        for (int j = 2; j <= tmp; j += 2) e[j].c++;
        dinic();
        for (int j = tmp + 1; j <= tot; j += 2) if (e[j].c) ans[i].push_back((j - tmp + 1) / 2);
    }
    rep(i, 0, minDeg) {
        printf("%d ", ans[i].size());
        for (auto y : ans[i]) printf("%d ", y);
        puts("");
    }
    return 0;
}

原文地址：https://www.cnblogs.com/aziint/p/9191660.html