「网络流 24 题」圆桌聚餐

所有餐桌连源点容量为餐桌容量，所有单位连汇点容量为单位人数，题目要求同一单位不能再同一餐桌就餐，那么对于每个餐桌，与所有单位建边且容量为1即可，最后求一遍最大流。

#include<stdio.h>#include<algorithm>#include<string.h>#include<queue>#include<vector>using namespace std;const int INF = 1e9 + 7;const int maxm = 1005;const int maxn = 100005;struct node{int v, flow, next;}edge[maxn];vector<int>p[maxm];int cur[maxn], head[maxn], vis[maxm], dis[maxm], pre[maxm], f[maxm][maxm];int cnt, s, t, n, m, rev = 0;void init(){rev = 0, cnt = 0, s = 0, t = n + m + 1;memset(head, -1, sizeof(head));}void add(int u, int v, int w){edge[cnt].v = v, edge[cnt].flow = w, edge[cnt].next = head[u], head[u] = cnt++;edge[cnt].v = u, edge[cnt].flow = 0, edge[cnt].next = head[v], head[v] = cnt++;}int bfs(){memset(pre, -1, sizeof(pre));memset(dis, -1, sizeof(dis));queue<int>q;dis[s] = 0;q.push(s);while (!q.empty()){int u = q.front();q.pop();for (int i = head[u];i != -1;i = edge[i].next){int v = edge[i].v;if (dis[v] == -1 && edge[i].flow){pre[v] = u;dis[v] = dis[u] + 1;q.push(v);}}}if (dis[t] == -1) return 0;return 1;}int dfs(int u, int flow){if (u == t) return flow;for (int i = cur[u];i != -1;i = edge[i].next){int v = edge[i].v;if (dis[v] == dis[u] + 1 && edge[i].flow){int d = dfs(v, min(edge[i].flow, flow));if (d > 0){edge[i].flow -= d, edge[i ^ 1].flow += 1;return d;}}}return 0;}int dinic(){int ans = 0, d;while (bfs()){for (int i = s;i <= t;i++) cur[i] = head[i];while (d = dfs(s, INF))ans += d;}return ans;}int main(){int i, j, k, sum = 0;scanf("%d%d", &n, &m);init();for (i = 1;i <= n;i++){scanf("%d", &k);sum += k;add(i + m, t, k);}for (i = 1;i <= m;i++){scanf("%d", &k);add(s, i, k);for (j = 1;j <= n;j++)add(i, j + m, 1);}if (sum != dinic()){printf("0\n");return 0;}printf("1\n");for (i = 1;i <= m;i++){for (j = head[i];j != -1;j = edge[j].next){if (edge[j].flow == 0){int v = edge[j].v;if (v > m&&v < t)p[v].push_back(i);}}}for (i = 1;i <= n;i++){for (j = 0;j < p[i + m].size();j++)printf("%d ", p[i + m][j]);printf("\n");}return 0;}