网络流24题（07）试题库问题（二分图多重匹配 + 最大流）

题意：

假设一个试题库中有 n 道试题。每道试题都标明了所属类别。同一道题可能有多个类别属性。

现要从题库中抽取 m 道题组成试卷。并要求试卷包含指定类型的试题。试设计一个满足要求的组卷算法。

思路：

1. 试卷和属性分别定义为 X, Y 集。每个试卷有多重属性，则由试卷分别向属性引弧，容量为 1，s 向 X 引弧容量为 1，Y 向 t 引弧，容量为需要的数量;

2. 求上面二分图的最大流即可。如果最大流等于需要选择出来的总试题数 m，则存在方案，输出即可；否而输出 No solution.

 

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <cstring>

#include <queue>

#include <vector>

using namespace std;



const int MAXN = 1050;

const int INFS = 0x3FFFFFFF;



struct edge {

    int from, to, cap, flow;

    edge(int _from, int _to, int _cap, int _flow) 

        : from(_from), to(_to), cap(_cap), flow(_flow) {}

};



class Dinic {

public:

    void initdata(int n, int s, int t) {

        this->n = n, this->s = s, this->t = t;

        edges.clear();

        for (int i = 0; i < n; i++)

            G[i].clear();

    }

    void addedge(int u, int v, int cap) {

        edges.push_back(edge(u, v, cap, 0));

        edges.push_back(edge(v, u, 0, 0));

        G[u].push_back(edges.size() - 2);

        G[v].push_back(edges.size() - 1);

    }

    bool BFS() {

        for (int i = 0; i < n; i++)

            vis[i] = false, d[i] = 0;

        queue<int> Q;

        Q.push(s);

        vis[s] = true;

        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 (e.cap > e.flow && !vis[e.to]) {

                    vis[e.to] = true;

                    d[e.to] = d[x] + 1;

                    Q.push(e.to);

                }

            }

        }

        return vis[t];

    }

    int DFS(int x, int aug) {

        if (x == t || aug == 0) return aug;

        int flow = 0;

        for (int i = 0; i < G[x].size(); i++) {

            edge& e = edges[G[x][i]];

            if (d[e.to] == d[x] + 1) {

                int f = DFS(e.to, min(aug, e.cap - e.flow));

                if (f <= 0) continue;

                e.flow += f;

                edges[G[x][i]^1].flow -= f;

                flow += f;

                aug -= f;

                if (aug == 0) break;

            } 

        }

        return flow;

    }

    int maxflow() {

        int flow = 0;

        while (BFS()) {

            flow += DFS(s, INFS);

        }

        return flow;

    }

    void getans(int p, vector<int>& ans) {

        ans.clear();

        for (int i = 0; i < G[p].size(); i++) {

            edge& e = edges[G[p][i]^1];

            if (e.to == p && e.flow == 1)

                ans.push_back(e.from);

        }

        sort(ans.begin(), ans.end());

    }

private:

    vector<edge> edges;

    vector<int> G[MAXN];

    int n, s, t, d[MAXN];

    bool vis[MAXN];

};



Dinic dc;



int main() {

    int k, n;

    scanf("%d%d", &k, &n);

    

    int s = 0, t = k + n + 1;

    dc.initdata(t + 1, s, t);



    int sum = 0;

    for (int i = 1; i <= k; i++) {

        int a;

        scanf("%d", &a);

        sum += a;

        dc.addedge(i + n, t, a);

    }

    for (int i = 1; i <= n; i++) {

        int p, a;

        scanf("%d", &p);

        for (int j = 0; j < p; j++) {

            scanf("%d", &a);

            dc.addedge(i, a + n, 1);

        }

        dc.addedge(s, i, 1);

    }

    int flow = dc.maxflow();

    if (flow == sum) {

        vector<int> ans;

        for (int i = 1; i <= k; i++) {

            dc.getans(i + n, ans);

            printf("%d:", i);

            for (int j = 0; j < ans.size(); j++)

                printf(" %d", ans[j]);

            printf("\n");

        }

    } else  printf("No Solution!\n");



    return 0;

}