POJ 3281 Dining (网络流之最大流)

时间：2024-07-11 00:07:20

题意：农夫为他的 N (1 ≤ N ≤ 100) 牛准备了 F (1 ≤ F ≤ 100)种食物和 D (1 ≤ D ≤ 100) 种饮料。每头牛都有各自喜欢的食物和饮料，

而每种食物或饮料只能分配给一头牛。最多能有多少头牛可以同时得到喜欢的食物和饮料？

析：是一个经典网络流的题，建立一个超级源点，连向每种食物，建立一个超级汇点，连向每种饮料，然后把每头牛拆成两个点，

一个和食物连，一个和饮料连，最后跑一遍最大流即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 400 + 5;

const int mod = 1e9;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

  return r >= 0 && r < n && c >= 0 && c < m;

}

struct Edge{

  int from, to, cap, flow;

};

struct Dinic{

  int n, m, s, t;

  vector<Edge> edges;

  vector<int> G[maxn];

  bool vis[maxn];

  int d[maxn];

  int cur[maxn];

  void init(){

    edges.clear();

    for(int i = 0; i < maxn; ++i)  G[i].clear();

  }

  void addEdge(int from, int to, int cap){

    edges.push_back((Edge){from, to, cap, 0});

    edges.push_back((Edge){to, from, 0, 0});

    m = edges.size();

    G[from].push_back(m-2);

    G[to].push_back(m-1);

  }

  bool bfs(){

    memset(vis, 0, sizeof vis);

    queue<int> q;

    q.push(s);

    d[s] = 0;

    vis[s] = 1;

    while(!q.empty()){

      int x = q.front();  q.pop();

      for(int i = 0; i < G[x].size(); ++i){

        Edge &e = edges[G[x][i]];

        if(!vis[e.to] && e.cap > e.flow){

          vis[e.to] = 1;

          d[e.to] = d[x] + 1;

          q.push(e.to);

        }

      }

    }

    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 < G[x].size(); ++i){

      Edge &e = edges[G[x][i]];

      if(d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){

        e.flow += f;

        edges[G[x][i]^1].flow -= f;

        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()){

      memset(cur, 0, sizeof cur);

      flow += dfs(s, INF);

    }

    return flow;

  }

};

Dinic dinic;

int main(){

  int k;

  while(scanf("%d %d %d", &n, &m, &k) == 3){

    dinic.init();

    int s = 0, t = 402;

    for(int i = 1; i <= m; ++i)  dinic.addEdge(s, i+200, 1);

    for(int i = 1; i <= k; ++i)  dinic.addEdge(i+300, t, 1);

    for(int i = 1; i <= n; ++i){

      int f, d;

      dinic.addEdge(i, i+100, 1);

      scanf("%d %d", &f, &d);

      for(int j = 0; j < f; ++j){

        int x;

        scanf("%d", &x);

        dinic.addEdge(x+200, i, 1);

      }

      for(int j = 0; j < d; ++j){

        int x;

        scanf("%d", &x);

        dinic.addEdge(i+100, x+300, 1);

      }

    }

    printf("%d\n", dinic.maxFlow(s, t));

  }

  return 0;

}

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