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;
}