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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。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 | /* ******************************************************************************** */#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;}/* ******************************************************************************** */ |
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原文地址：http://www.cnblogs.com/jklongint/p/4695002.html
