luogu P3254 圆桌问题

简单的最大流问题，一样的设源点汇点，每个单位向每个餐桌连capacity为1的边，源点向每个单位连capacity为人数的边，每个餐桌向汇点连capacity为座位数的边，跑最大流，若flow=总人数则有方案，遍历每一个单位流量为1的点即可

#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;

const int maxm = 1e5+5;
const int INF = 0x3f3f3f3f;

struct edge{
    int u, v, cap, flow, nex;
} edges[maxm];

int head[maxm], cur[maxm], cnt, level[1024];

void init() {
    memset(head, -1, sizeof(head));
}

void add(int u, int v, int cap) {
    edges[cnt] = edge{u, v, cap, 0, head[u]};
    head[u] = cnt++;
}

void addedge(int u, int v, int cap) {
    add(u, v, cap), add(v, u, 0);
}

void bfs(int s) {
    memset(level, -1, sizeof(level));
    queue<int> q;
    level[s] = 0;
    q.push(s);
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        for(int i = head[u]; i != -1; i = edges[i].nex) {
            edge& now = edges[i];
            if(now.cap > now.flow && level[now.v] < 0) {
                level[now.v] = level[u] + 1;
                q.push(now.v);
            }
        }
    }
}

int dfs(int u, int t, int f) {
    if(u == t) return f;
    for(int& i = cur[u]; i != -1; i = edges[i].nex) {
        edge& now = edges[i];
        if(now.cap > now.flow && level[u] < level[now.v]) {
            int d = dfs(now.v, t, min(f, now.cap - now.flow));
            if(d > 0) {
                now.flow += d;
                edges[i^1].flow -= d;
                return d;
            }

        }
    }
    return 0;
}

int dinic(int s, int t) {
    int maxflow = 0;
    for(;;) {
        bfs(s);
        if(level[t] < 0) break;
        memcpy(cur, head, sizeof(head));
        int f;
        while((f = dfs(s, t, INF)) > 0)
            maxflow += f;
    }
    return maxflow;
}

void run_case() {
    int n, m, cap, sum = 0;
    init();
    cin >> m >> n;
    int s = 0, t = m+n+1;
    for(int i = 1; i <= m; ++i) {
        cin >> cap;
        sum += cap;
        addedge(s, i, cap);
    }
    for(int i = 1; i <= n; ++i) {
        cin >> cap;
        addedge(i+m, t, cap);
    }
    for(int i = 1; i <= m; ++i)
        for(int j = 1; j <= n; ++j)
            addedge(i, j+m, 1);
    int flow = dinic(s, t);
    if(flow != sum) {
        cout << 0;
        return;
    } else {
        cout << "1\n";
        for(int i = 1; i <= m; ++i) {
            for(int u = head[i]; u!=-1; u = edges[u].nex) {
                int v = edges[u].v;
                if(v != s && edges[u].flow) {
                    edges[u].flow--;
                    cout << v-m << " ";
                }
            }
            cout << "\n";
        }
    }
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    run_case();
    cout.flush();
    return 0;
}

 

来源：https://www.cnblogs.com/GRedComeT/p/12298506.html