[hdu2119]二分图最小覆盖，最大匹配

题意：给一个01矩阵，每次可以选一行或一列，打掉上面所有的1，求打掉所有的1所需的最小次数。

思路：经典的模型了，二分图最小覆盖=最大匹配。所谓最小覆盖是指选最少的点关联所有的边。容易得到将行和列看成点，1看成边，那么就是选尽量少的行和列来关联所有的1，最小覆盖模型，用最大匹配做。可以选择匈牙利算法，或者直接最大流。

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/* ******************************************************************************** */ #include <iostream>                                                                 // #include <cstdio>                                                                   // #include <cmath>                                                                    // #include <cstdlib>                                                                  // #include <cstring>                                                                  // #include <vector>                                                                   // #include <ctime>                                                                    // #include <deque>                                                                    // #include <queue>                                                                    // #include <algorithm>                                                                // #include <map>                                                                      // #include <cmath>                                                                    // using namespace std;                                                                //                                                                                     // #define pb push_back                                                                // #define mp make_pair                                                                // #define X first                                                                     // #define Y second                                                                    // #define all(a) (a).begin(), (a).end()                                               // #define fillchar(a, x) memset(a, x, sizeof(a))                                      //                                                                                     // typedef pair<int, int> pii;                                                         // typedef long long ll;                                                               // typedef unsigned long long ull;                                                     //                                                                                     // #ifndef ONLINE_JUDGE                                                                // void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}    // void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>                    // void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;          // while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>      // void print(const T t){cout<<t<<endl;}template<typename F,typename...R>              // void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>   // void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}   // #endif // ONLINE_JUDGE                                                              // template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}        // template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}        // template<typename T>                                                                // void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];}            // template<typename T>                                                                // void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];}            //                                                                                     // const double PI = acos(-1.0);                                                       // const int INF = 1e9 + 7;                                                            //                                                                                     // /* -------------------------------------------------------------------------------- */   struct Edmonds {     const static int maxn = 1e2 + 7;     int n, m;     bool g[maxn][maxn];     bool vis[maxn];     int left[maxn];       void init(int n, int m) {         this->n = n;         this->m = m;         memset(g, 0, sizeof(g));         memset(left, -1, sizeof(left));     }     void add(int u, int v) {         g[u][v] = true;     }     bool match(int u) {         for(int v = 1; v <= m; v++)if(g[u][v] && !vis[v]) {                 vis[v] = true;                 if(left[v] == -1 || match(left[v])) {                     left[v] = u;                     return true;                 }             }         return false;     }       int solve() {         int ans = 0;         for(int i = 1; i <= n; i++) {             memset(vis, 0, sizeof(vis));             if(match(i)) ans++;         }         return ans;     } };/** 点从1开始编号 **/ Edmonds solver;   int main() { #ifndef ONLINE_JUDGE     freopen("in.txt", "r", stdin); #endif // ONLINE_JUDGE     int n, m;     while (cin >> n, n) {         cin >> m;         solver.init(n, m);         for (int i = 0; i < n; i ++) {             for (int j = 0; j < m; j ++) {                 int x;                 scanf("%d", &x);                 if (x) solver.add(i + 1, j + 1);             }         }         cout << solver.solve() << endl;     }     return 0; } /* ******************************************************************************** */