[题解]poj 1274 The Perfect Stall(网络流)

二分匹配传送门[here]

原题传送门[here]

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　　题意大概说一下，就是有N头牛和M个牛棚，每头牛愿意住在一些牛棚，求最大能够满足多少头牛的要求。

　　很明显就是一道裸裸的二分图最大匹配，但是为了练练网络流（做其它的题的时候，神奇地re掉了，于是就写基础题了）的最大流算法，就做做这道题。

　　每一个牛都可一看成是个源点，每一个牛棚都可以看成是个汇点，但是一个网络应该只有一个汇点和一个源点才对，于是构造一个连接每个牛的超级源点，一个连接每个牛棚的超级汇点，每条边的容量为1，然后最大流Dinic算法（其它最大流算法也行）就行了。

Code_{极其不简洁的代码}

 /**
  * poj.org
  * Problem#1274
  * Accepted
  * Time:16ms
  * Memory:1980k
  */
 #include<iostream>
 #include<sstream>
 #include<cstdio>
 #include<cmath>
 #include<cstdlib>
 #include<cstring>
 #include<cctype>
 #include<ctime>
 #include<queue>
 #include<set>
 #include<map>
 #include<stack>
 #include<vector>
 #include<algorithm>
 using namespace std;
 typedef bool boolean;
 #define smin(a, b) (a) = min((a), (b))
 #define smax(a, b) (a) = max((a), (b))
 #define INF 0xfffffff
 template<typename T>
 inline boolean readInteger(T& u){
     char x;
     int aFlag = ;
     while(!isdigit((x = getchar())) && x != '-' && ~x);
     if(!(~x))    return false;
     if(x == '-'){
         aFlag = -;
         x = getchar();
     }
     for(u = x - ''; isdigit((x = getchar())); u = u *  + x - '');
     ungetc(x, stdin);
     u *= aFlag;
     return true;
 }
 ///map template starts
 typedef class Edge{
     public:
         int end;
         int next;
         int cap;
         int flow;
         Edge(const int end = , const int next = , const int cap = , const int flow = ):end(end), next(next), cap(cap), flow(flow){}
 }Edge;

 typedef class MapManager{
     public:
         int ce;
         int *h;
         Edge *edge;
         MapManager(){}
         MapManager(int points, int limit):ce(){
             h = new int[(const int)(points + )];
             edge = new Edge[(const int)(limit + )];
             memset(h, , sizeof(int) * (points + ));
         }
         inline void addEdge(int from, int end, int cap, int flow){
             edge[++ce] = Edge(end, h[from], cap, flow);
             h[from] = ce;
         }
         inline void addDoubleEdge(int from, int end, int cap){
             addEdge(from, end, cap, );
             addEdge(end, from, cap, cap);
         }
         Edge& operator [](int pos){
             return edge[pos];
         }
         inline int reverse(int pos){        //反向边
             return (pos & ) ? (pos + ) : (pos - );
         }
         inline void clear(){
             delete[] edge;
             delete h;
             ce = ;
         }
 }MapManager;

 #define m_begin(g, i)     (g).h[(i)]
 #define m_end(g, i)     (g).edge[(i)].end
 #define m_next(g, i)     (g).edge[(i)].next
 #define m_cap(g, i)     (g).edge[(i)].cap
 #define m_flow(g, i)     (g).edge[(i)].flow
 ///map template ends

 int n, m;
 int t;
 MapManager g;

 inline boolean init(){
     if(!readInteger(n))    return false;
     readInteger(m);
     t = n + m + ;
     g = MapManager(t + , (n + ) * (m + ) * );
     for(int i = , a, b; i <= n; i++){
         readInteger(a);
         while(a--){
             readInteger(b);
             g.addDoubleEdge(i, b + n, );
         }
     }
     for(int i = ; i <= n; i++)
         g.addDoubleEdge(, i, );
     for(int i = ; i <= m; i++)
         g.addDoubleEdge(i + n, t, );
     return true;
 }

 boolean *visited;
 int* divs;
 queue<int> que;

 inline boolean getDivs(){
     memset(visited, false, sizeof(boolean) * (t + ));
     divs[] = ;
     visited[] = true;
     que.push();
     while(!que.empty()){
         int e = que.front();
         que.pop();
         for(int i = m_begin(g, e); i != ; i = g[i].next){
             int& eu = g[i].end;
             if(!visited[eu] && g[i].flow < g[i].cap){
                 divs[eu] = divs[e] + ;
                 visited[eu] = true;
                 que.push(eu);
             }
         }
     }
     return visited[t];
 }

 int blockedflow(int node, int minf){
     if(node == t || minf == )    return minf;
     int f, flow = ;
     for(int i = m_begin(g, node); i != ; i = m_next(g, i)){
         int& e = g[i].end;
         if(divs[e] == divs[node] +  && visited[e] && (f = blockedflow(e, min(minf, g[i].cap - g[i].flow))) > ){
             flow += f;
             g[i].flow += f;
             g[g.reverse(i)].flow -= f;
             minf -= f;
             if(minf == )    return flow;
         }
     }
     return flow;
 }

 inline int maxflow(){
     visited = new boolean[(const int)(t + )];
     divs = new int[(const int)(t + )];
     int res = ;
     while(getDivs()){
         res += blockedflow(, INF);
     }
     return res;
 }

 inline void solve(){
     int res = maxflow();
     cout << res << endl;
 }

 inline void clear(){
     delete[] visited;
     delete[] divs;
     g.clear();
 }

 int main(){
     while(init()){
         solve();
         clear();
     }
     return ;
 }