二分匹配传送门[here]
原题传送门[here]
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
1 /** 2 * poj.org 3 * Problem#1274 4 * Accepted 5 * Time:16ms 6 * Memory:1980k 7 */ 8 #include<iostream> 9 #include<sstream> 10 #include<cstdio> 11 #include<cmath> 12 #include<cstdlib> 13 #include<cstring> 14 #include<cctype> 15 #include<ctime> 16 #include<queue> 17 #include<set> 18 #include<map> 19 #include<stack> 20 #include<vector> 21 #include<algorithm> 22 using namespace std; 23 typedef bool boolean; 24 #define smin(a, b) (a) = min((a), (b)) 25 #define smax(a, b) (a) = max((a), (b)) 26 #define INF 0xfffffff 27 template<typename T> 28 inline boolean readInteger(T& u){ 29 char x; 30 int aFlag = 1; 31 while(!isdigit((x = getchar())) && x != '-' && ~x); 32 if(!(~x)) return false; 33 if(x == '-'){ 34 aFlag = -1; 35 x = getchar(); 36 } 37 for(u = x - '0'; isdigit((x = getchar())); u = u * 10 + x - '0'); 38 ungetc(x, stdin); 39 u *= aFlag; 40 return true; 41 } 42 ///map template starts 43 typedef class Edge{ 44 public: 45 int end; 46 int next; 47 int cap; 48 int flow; 49 Edge(const int end = 0, const int next = 0, const int cap = 0, const int flow = 0):end(end), next(next), cap(cap), flow(flow){} 50 }Edge; 51 52 typedef class MapManager{ 53 public: 54 int ce; 55 int *h; 56 Edge *edge; 57 MapManager(){} 58 MapManager(int points, int limit):ce(0){ 59 h = new int[(const int)(points + 1)]; 60 edge = new Edge[(const int)(limit + 1)]; 61 memset(h, 0, sizeof(int) * (points + 1)); 62 } 63 inline void addEdge(int from, int end, int cap, int flow){ 64 edge[++ce] = Edge(end, h[from], cap, flow); 65 h[from] = ce; 66 } 67 inline void addDoubleEdge(int from, int end, int cap){ 68 addEdge(from, end, cap, 0); 69 addEdge(end, from, cap, cap); 70 } 71 Edge& operator [](int pos){ 72 return edge[pos]; 73 } 74 inline int reverse(int pos){ //反向边 75 return (pos & 1) ? (pos + 1) : (pos - 1); 76 } 77 inline void clear(){ 78 delete[] edge; 79 delete h; 80 ce = 0; 81 } 82 }MapManager; 83 84 #define m_begin(g, i) (g).h[(i)] 85 #define m_end(g, i) (g).edge[(i)].end 86 #define m_next(g, i) (g).edge[(i)].next 87 #define m_cap(g, i) (g).edge[(i)].cap 88 #define m_flow(g, i) (g).edge[(i)].flow 89 ///map template ends 90 91 int n, m; 92 int t; 93 MapManager g; 94 95 inline boolean init(){ 96 if(!readInteger(n)) return false; 97 readInteger(m); 98 t = n + m + 1; 99 g = MapManager(t + 1, (n + 1) * (m + 1) * 2); 100 for(int i = 1, a, b; i <= n; i++){ 101 readInteger(a); 102 while(a--){ 103 readInteger(b); 104 g.addDoubleEdge(i, b + n, 1); 105 } 106 } 107 for(int i = 1; i <= n; i++) 108 g.addDoubleEdge(0, i, 1); 109 for(int i = 1; i <= m; i++) 110 g.addDoubleEdge(i + n, t, 1); 111 return true; 112 } 113 114 boolean *visited; 115 int* divs; 116 queue<int> que; 117 118 inline boolean getDivs(){ 119 memset(visited, false, sizeof(boolean) * (t + 1)); 120 divs[0] = 1; 121 visited[0] = true; 122 que.push(0); 123 while(!que.empty()){ 124 int e = que.front(); 125 que.pop(); 126 for(int i = m_begin(g, e); i != 0; i = g[i].next){ 127 int& eu = g[i].end; 128 if(!visited[eu] && g[i].flow < g[i].cap){ 129 divs[eu] = divs[e] + 1; 130 visited[eu] = true; 131 que.push(eu); 132 } 133 } 134 } 135 return visited[t]; 136 } 137 138 int blockedflow(int node, int minf){ 139 if(node == t || minf == 0) return minf; 140 int f, flow = 0; 141 for(int i = m_begin(g, node); i != 0; i = m_next(g, i)){ 142 int& e = g[i].end; 143 if(divs[e] == divs[node] + 1 && visited[e] && (f = blockedflow(e, min(minf, g[i].cap - g[i].flow))) > 0){ 144 flow += f; 145 g[i].flow += f; 146 g[g.reverse(i)].flow -= f; 147 minf -= f; 148 if(minf == 0) return flow; 149 } 150 } 151 return flow; 152 } 153 154 inline int maxflow(){ 155 visited = new boolean[(const int)(t + 1)]; 156 divs = new int[(const int)(t + 1)]; 157 int res = 0; 158 while(getDivs()){ 159 res += blockedflow(0, INF); 160 } 161 return res; 162 } 163 164 inline void solve(){ 165 int res = maxflow(); 166 cout << res << endl; 167 } 168 169 inline void clear(){ 170 delete[] visited; 171 delete[] divs; 172 g.clear(); 173 } 174 175 int main(){ 176 while(init()){ 177 solve(); 178 clear(); 179 } 180 return 0; 181 }