「网络流 24 题」试题库

思路

建立超级源点 $S$ 和超级汇点 $T$，将每一类题目拆分为入点 $in$ 和出点 $out$，连边 $in\to out$，边权为 $k_i$，也就是所需的题目数量，同时连边 $out\to T$；并对每一道题目及其对应的类型 $p_i$，连边 $i\to i{n}_{p_i}$，容量为 $1$，$S\to i$ 边权同样为 $1$

跑最大流即可，看看最大流量是否为 $m$，按照残余网络的流量来输出方案

#include<bits/stdc++.h>
#define fore(i,l,r)	for(int i=(int)(l);i<(int)(r);++i)
#define fi first
#define se second
#define endl '\n'
#define ull unsigned long long
#define ALL(v) v.begin(), v.end()
#define Debug(x, ed) std::cerr << #x << " = " << x << ed;

const int INF=0x3f3f3f3f;
const long long INFLL=1e18;

typedef long long ll;

constexpr int inf = 1E9;

template<class T>
struct Dinic {
    struct _Edge {
        int to;
        T cap;
        _Edge(int to, T cap) : to(to), cap(cap) {}
    };
    
    int n; //点的数量，编号从 1 开始
    std::vector<_Edge> e; //链式前向星
    std::vector<std::vector<int>> g; //起到链式前向星nxt的作用
    std::vector<int> cur; //当前弧优化
    std::vector<int> h; //深度
    
    Dinic() {}
    Dinic(int n) {
        init(n);
    }
    
    void init(int n) {
        this->n = n;
        e.clear();
        g.assign(n + 1, {});
        cur.resize(n + 1);
        h.resize(n + 1);
    }
    
    bool bfs(int s, int t) { //构造分层图
        h.assign(n + 1, -1);
        std::queue<int> que;
        h[s] = 0;
        que.push(s);
        while (!que.empty()) {
            const int u = que.front();
            que.pop();
            for (int i : g[u]) {
                auto [v, c] = e[i];
                if (c > 0 && h[v] == -1) { //下一层有容量的邻居
                    h[v] = h[u] + 1;
                    if (v == t) {
                        return true;
                    }
                    que.push(v);
                }
            }
        }
        return false;
    }
    
    T dfs(int u, int t, T f) {
        if (u == t) {
            return f;
        }
        auto r = f;
        for (int &i = cur[u]; i < int(g[u].size()); ++i) {
            const int j = g[u][i];
            auto [v, c] = e[j];
            if (c > 0 && h[v] == h[u] + 1) {
                auto a = dfs(v, t, std::min(r, c));
                e[j].cap -= a;
                e[j ^ 1].cap += a;
                r -= a; //r是剩余可用流量
                if (r == 0) {
                    return f;  //如果r用完，说明f跑满了
                }
            }
        }
        return f - r; //否则f-r就是已用流量
    }
    void addEdge(int u, int v, T c) {
        g[u].push_back(e.size()); //记录在e中的下标
        e.emplace_back(v, c);
        g[v].push_back(e.size()); //反向边
        e.emplace_back(u, 0);
    }
    T flow(int s, int t) {
        T ans = 0;
        while (bfs(s, t)) {
            cur.assign(n + 1, 0); //当前弧初始化
            ans += dfs(s, t, std::numeric_limits<T>::max());
        }
        return ans;
    }
    
    std::vector<bool> minCut() { //最小割
        std::vector<bool> c(n + 1);
        for (int i = 1; i <= n; i++) {
            c[i] = (h[i] != -1);
        }
        return c;
    }
    
    struct Edge {
        int from;
        int to;
        T cap;
        T flow;
    };
    std::vector<Edge> edges() {
        std::vector<Edge> a;
        for (int i = 0; i < e.size(); i += 2) {
            Edge x;
            x.from = e[i + 1].to;
            x.to = e[i].to;
            x.cap = e[i].cap + e[i + 1].cap;
            x.flow = e[i + 1].cap;
            a.push_back(x);
        }
        return a;
    }
};

int main(){
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    int k, n;
    std::cin >> k >> n;
    int sum = 0;
    int S = 2 * n + k + 3, T = S + 1;
    Dinic<int> dinic(2 * n + k + 5);
    fore(i, 1, k + 1){
        int w;
        std::cin >>  w;
        sum += w;
        dinic.addEdge(2 * n + i, T, w);
    }
    fore(i, 1, n + 1){
        int in = 2 * i - 1, out = 2 * i;
        dinic.addEdge(S, in, INF);
        dinic.addEdge(in, out, 1);
        int m;
        std::cin >> m;
        while(m--){
            int x;
            std::cin >> x;
            dinic.addEdge(out, 2 * n + x, 1);
        }
    }

    if(dinic.flow(S, T) != sum){
        std::cout << "No Solution!";
        return 0;
    }


    auto edge = dinic.edges();
    std::vector<std::vector<int>> ans(k + 1, std::vector<int>());
    for(auto [fr, to, cap, flow] : edge){
        if(fr <= 2 * n && fr % 2 == 0 && to > 2 * n && to <= 2 * n + k && flow){
            int u = fr >> 1, v = to - 2 * n;
            ans[v].push_back(u);
        }
    }

    fore(i, 1, k + 1){
        std::cout << i << ": ";
        for(auto x : ans[i]) std::cout << x << ' ';
        std::cout << endl;
    }

    return 0;
}