网络流24题（04）圆桌问题（二分图多重匹配 + 最大流）

思路：

1. 对于任何代表 Xi，他可以坐在任意一张桌子 Yi 上，抽象出来这个条件就可以方便下面我们建模了;

2. 从源点向 Xi 引弧，容量为代表数目。从 Yi 向汇点引弧，容量为座子所能容纳的人数。从 Xi 分别向每个 Yi 引弧，容量为 1，表示代表对每张桌子都有选择权;

3. 求二分图的最大流即可，如果代表的总数目 = 最大流，则表示每个代表都能找到自己的位置，题目有解，否则无解。

 

#include <iostream>

#include <algorithm>

#include <queue>

#include <vector>

using namespace std;



const int MAXN = 500;

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

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

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

        int m = edges.size();

        G[u].push_back(m - 2);

        G[v].push_back(m - 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 (!vis[e.to] && e.cap > e.flow) {

                    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 s, int t) {

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

        int flow = 0;

        while (BFS()) {

            flow += DFS(s, INFS);

        }

        return flow;

    }

    void cleardata(int n) {

        this->n = n;

        edges.clear();

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

            G[i].clear();

    }

    void print(int m) {

        for (int x = 1; x <= m; x++) {

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

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

                if (e.cap > 0 && e.cap == e.flow && e.to != t)

                    printf("%d ", e.to - m);

            }

            printf("\n");

        }

    }

private:

    vector<edge> edges;

    vector<int> G[MAXN];

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

    bool vis[MAXN];

};



Dinic dc;



int main() {

    int m, n;

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



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

    dc.cleardata(t + 1);



    int sum = 0;

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

        int a;

        scanf("%d", &a);

        sum += a;

        dc.addedge(s, i, a);

    }

    

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

        int a;

        scanf("%d", &a);

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

    }

    for (int i = 1; i <= m; i++)

        for (int j = 1; j <= n; j++)

            dc.addedge(i, j + m, 1);



    int f = dc.maxflow(s, t);

    if (f == sum) {

        printf("1\n");

        dc.print(m);

    } else {

        printf("0\n");

    }



    return 0;

}