[题目链接]
http://poj.org/problem?id=1325
[算法]
二分图最小覆盖
[代码]
#include <algorithm> #include <bitset> #include <cctype> #include <cerrno> #include <clocale> #include <cmath> #include <complex> #include <cstdio> #include <cstdlib> #include <cstring> #include <ctime> #include <deque> #include <exception> #include <fstream> #include <functional> #include <limits> #include <list> #include <map> #include <iomanip> #include <ios> #include <iosfwd> #include <iostream> #include <istream> #include <ostream> #include <queue> #include <set> #include <sstream> #include <stdexcept> #include <streambuf> #include <string> #include <utility> #include <vector> #include <cwchar> #include <cwctype> #include <stack> #include <limits.h> using namespace std; #define MAXN 110 #define MAXK 1010 struct edge { int to,nxt; } e[MAXK]; int i,n,m,k,ans,tot,t,x,y; int match[MAXN << 1],head[MAXN << 1]; bool visited[MAXN << 1]; inline void addedge(int u,int v) { tot++; e[tot] = (edge){v,head[u]}; head[u] = tot; } inline bool hungary(int u) { int i,v; visited[u] = true; for (i = head[u]; i; i = e[i].nxt) { v = e[i].to; if (!visited[v]) { visited[v] = true; if (!match[v] || hungary(match[v])) { match[v] = u; return true; } } } return false; } int main() { while (scanf("%d",&n) && n) { tot = 0; memset(head,0,sizeof(head)); memset(match,0,sizeof(match)); scanf("%d%d",&m,&k); for (i = 1; i <= k; i++) { scanf("%d%d%d",&t,&x,&y); if (x * y) addedge(x,y + n); } ans = 0; for (i = 1; i <= n; i++) { memset(visited,false,sizeof(visited)); if (hungary(i)) ans++; } printf("%d\n",ans); } return 0; }