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;
}