「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;
}