【网络流24题】圆桌聚餐 （最大流）

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题意

假设有来自 $n$ 个不同单位的代表参加一次国际会议。每个单位的代表数分别为 $r_i$​​。会议餐厅共有 $m$ 张餐桌，每张餐桌可容纳 $c_i$​​ 个代表就餐。
为了使代表们充分交流，希望从同一个单位来的代表不在同一个餐桌就餐。

试设计一个算法，给出满足要求的代表就餐方案。

题解

最大流

代码

#include<bits/stdc++.h>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int nmax = 1e6+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const ull p = 67;
const ull MOD = 1610612741;
int n, m;
int ss[nmax], top;
struct Dinic {
    int head[nmax], cur[nmax], d[nmax];
    bool vis[nmax];
    int tot, n, m, s, t, front, tail;
    int qqq[nmax];
    struct edge {
        int nxt, to, w, cap, flow;
    } e[nmax<<1];
    void init(int n) {
        this->n = n;
        memset(head, -1, sizeof head);
        memset(e,0,sizeof e);
        this->tot = 0;
    }
    int add_edge(int u, int v, int c) {
        int temp = tot;
        e[tot].to = v, e[tot].cap = c, e[tot].flow = 0;
        e[tot].nxt = head[u];
        head[u] = tot++;
        e[tot].to = u, e[tot].cap = c, e[tot].flow = c;
        e[tot].nxt = head[v];
        head[v] = tot++;
        return temp;
    }
    bool BFS() {
//        memset(vis, 0, sizeof(vis));
//        queue<int>Q;
        for(int i = 0; i <= n; ++i) vis[i] = false;
        front = tail = 0;
        vis[s] = 1; d[s] = 0;
//        Q.push(s);
        qqq[tail++] = s;
        while (front < tail) {
//            int u = Q.front(); Q.pop();
            int u = qqq[front++];
            for (int i = head[u]; i != -1; i = e[i].nxt) {
                int v = e[i].to;
                if (!vis[v] && e[i].cap > e[i].flow) {
                    vis[v] = 1;
                    d[v] = d[u] + 1;
//                    Q.push(v);
                    qqq[tail++] = v;
                }
            }
        }
        return vis[t];
    }
    int DFS(int x, int a) {
        if (x == t || a == 0) return a;
        int Flow = 0, f;
        for (int& i = cur[x]; i != -1; i = e[i].nxt) {
            int v = e[i].to;
            if (d[v] == d[x] + 1 && (f = DFS(v, min(a, e[i].cap - e[i].flow))) > 0) {
                Flow += f;
                e[i].flow += f;
                e[i ^ 1].flow -= f;
                a -= f;
                if (a == 0) break;
            }
        }
        return Flow;
    }
    int Maxflow(int s, int t) {
        this->s = s, this->t = t;
        int Flow = 0;
        while (BFS()) {
            for (int i = 0; i <= n; i++) cur[i] = head[i];
            Flow += DFS(s,INF);
        }
        return Flow;
    }
} dinic;
int main(){
    scanf("%d %d", &n, &m);
    int s = 0, t = n + m + 1, tmp;
    dinic.init(t);
    int totpeo = 0;
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &tmp);
        totpeo  += tmp;
        dinic.add_edge(s, i, tmp);
        for(int j = n + 1; j <= n + m; ++j) {
            dinic.add_edge(i, j, 1);
        }
    }
    for(int j = n + 1; j <= n + m; ++j) {
        scanf("%d", &tmp);
        dinic.add_edge(j, t, tmp);
    }
    int mxflow = dinic.Maxflow(s, t);
    if(mxflow == totpeo) {
        printf("1\n");
        for(int u = 1; u <= n; ++u) {
            top = 0;
            for(int i = dinic.head[u]; i != -1; i = dinic.e[i].nxt) {
                int v = dinic.e[i].to;
                if(i % 2 == 0 && dinic.e[i].flow == 1) {
                    ss[++top] = v - n;
                }
            }
            for(int i = 1; i <= top; ++i) {
                if(i == 1) printf("%d", ss[i]);
                else printf(" %d", ss[i]);
            }
            printf("\n");
        }
    } else {
        printf("0\n");
    }
    return 0;
}