以行列为点建图,每个点(x,y) 对应一条边连接x,y。二分图的最小点覆盖=最大匹配
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; #define pb(a) push(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c)); } template<class T> T max(const T& a,const T& b,const T& c) { return max(max(a,b),max(a,c)); } void debug() { #ifdef ONLINE_JUDGE #else freopen("data.in","r",stdin); // freopen("d:\\out1.txt","w",stdout); #endif } int getch() { int ch; while((ch=getchar())!=EOF) { if(ch!=‘ ‘&&ch!=‘\n‘)return ch; } return EOF; } const int maxn = 1100; vector<int> g[maxn]; int n, m; void Init() { for(int i=1; i<=n; i++) g[i].clear(); } void add(int u, int v) { g[u].push_back(v); } int match[maxn]; int vis[maxn]; bool dfs(int u) { vis[u] = true; for(int i = 0; i < g[u].size(); i++) { int v = g[u][i]; int w = match[v]; if(w<0||!vis[w]&&dfs(w)) { match[u] = v; match[v] = u; return true; } } return false; } int solve() { memset(match,-1,sizeof(match)); int res = 0; for(int u=1; u <= n; u++) { if(match[u]<0) { memset(vis, 0, sizeof(vis)); if(dfs(u)) res++; } } return res; } int main() { debug(); while(scanf("%d%d", &n, &m) != EOF) { for(int i=1; i<=m; i++) { int u,v; scanf("%d%d", &u, &v); add(u, v+n); add(v+n, u); } printf("%d\n",solve()); } return 0; }
POJ 3041 Asteroids 二分图匹配,布布扣,bubuko.com
原文:http://www.cnblogs.com/BMan/p/3713655.html