【最大流/二分图匹配】【网络流24题】【P3254】 圆桌问题

Description

假设有来自m 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ri (i =1,2,……,m)。

会议餐厅共有n 张餐桌，每张餐桌可容纳ci (i =1,2,……,n)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。

对于给定的代表数和餐桌数以及餐桌容量，编程计算满足要求的代表就餐方案。输出方案。

Limitation

\(1~\leq~m~\leq~50,~1~\leq~n~\leq~270\)

Solution

好裸的最大流啊……网络流24题怎么有这么水的题……

源点向单位连边，容量为人数

单位向餐桌连边，容量为 \(1\)，代表一个餐桌最多做一个该单位的人

餐桌向汇点连边，容量为餐桌大小，代表最多做多少人

跑个最大流就好了……

输出方案的话，对每个单位，枚举看向哪个餐桌的流量为 \(1\) 即可

貌似还可以神仙贪心？

Code

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#ifdef ONLINE_JUDGE
#define freopen(a, b, c)
#endif

typedef long long int ll;

namespace IPT {
  const int L = 1000000;
  char buf[L], *front=buf, *end=buf;
  char GetChar() {
    if (front == end) {
      end = buf + fread(front = buf, 1, L, stdin);
      if (front == end) return -1;
    }
    return *(front++);
  }
}

template <typename T>
inline void qr(T &x) {
  char ch = IPT::GetChar(), lst = ' ';
  while ((ch > '9') || (ch < '0')) lst = ch, ch=IPT::GetChar();
  while ((ch >= '0') && (ch <= '9')) x = (x << 1) + (x << 3) + (ch ^ 48), ch = IPT::GetChar();
  if (lst == '-') x = -x;
}

namespace OPT {
  char buf[120];
}

template <typename T>
inline void qw(T x, const char aft, const bool pt) {
  if (x < 0) {x = -x, putchar('-');}
  int top=0;
  do {OPT::buf[++top] = static_cast<char>(x % 10 + '0');} while (x /= 10);
  while (top) putchar(OPT::buf[top--]);
  if (pt) putchar(aft);
}

const int maxn = 400;
const int INF = 0x3f3f3f3f;

struct Edge {
  int u, v, flow;
  Edge *nxt, *bk;

  Edge(const int _u, const int _v, const int _flow, Edge* &h)
      : u(_u), v(_v), flow(_flow), nxt(h) {
    h = this;
  }
};
Edge *hd[maxn], *fir[maxn];
inline void cont(const int _u, const int _v, const int _flow) {
  auto u = new Edge(_u, _v, _flow, hd[_u]), v = new Edge(_v, _u, 0, hd[_v]);
  (u->bk = v)->bk = u;
}

int n, m, s, t, ans;
int MU[maxn], CU[maxn], lp[maxn], rp[maxn], dist[maxn];
std::queue<int>Q;

bool bfs();
int dfs(const int u, int canag);
void printans();

int main() {
  freopen("1.in", "r", stdin);
  qr(n); qr(m);
  for (int i = 1; i <= n; ++i) {
    qr(MU[i]);
    lp[i] = ++t;
  }
  for (int i = 1; i <= m; ++i) {
    qr(CU[i]); rp[i] = ++t;
  }
  s = ++t; ++t;
  for (int i = 1; i <= n; ++i) {
    cont(s, lp[i], MU[i]);
  }
  for (int i = 1; i <= m; ++i) {
    cont(rp[i], t, CU[i]);
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      cont(lp[i], rp[j], 1);
    }
  }
  while (bfs()) {
    for (int i = 1; i <= t; ++i) fir[i] = hd[i];
    dfs(s, INF);
  }
  printans();
  return 0;
}

void printans() {
  ans = 1;
  for (auto e = hd[s]; e; e = e->nxt) {
    ans &= (e->flow == 0);
  }
  qw(ans, '\n', true);
  if (ans) {
    for (int u = 1; u <= n; ++u) {
      for (auto e = hd[u]; e; e = e->nxt) if (e->flow == 0) {
        qw(e->v - n, ' ', true);
      }
      putchar('\n');
    }
  }
}

bool bfs() {
  memset(dist, 0, sizeof dist);
  dist[s] = 1; Q.push(s);
  while (!Q.empty()) {
    int u = Q.front(); Q.pop();
    for (auto e = hd[u]; e; e = e->nxt) if (e->flow > 0) {
      int v = e->v;
      if (dist[v] == 0) {
        dist[v] = dist[u] + 1;
        Q.push(v);
      }
    }
  }
  return dist[t];
}

int dfs(const int u, int canag) {
  if ((u == t) || (!canag)) return canag;
  int _f = 0;
  for (auto &e = fir[u]; e; e = e->nxt) if (e->flow > 0) {
    int v = e->v;
    if (dist[v] == (dist[u] + 1)) {
      int f = dfs(v, std::min(e->flow, canag));
      e->flow -= f; e->bk->flow += f; _f += f;
      if (!(canag -= f)) {
        break;
      }
    }
  }
  return _f;
}