- 嗯
想不通 - 就是二分之后的点，寻找左边的点和右边的点的保证两条边的顶点不相同的最大边数
匈牙利算法 O(mn)
左边寻找和右边相邻的边，如果右边还没有和左边进行连线，那么匹配成功。如果右边已经进行连线，那么考虑左边是否能更改连线，换一个右边。
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;public class Main {private static int N = 510;private static int M = 100010;private static int[] h = new int[M];private static int[] e = new int[M];private static int[] ne = new int[M];private static int n1,n2,m;private static int[] match = new int[N];private static boolean[] vis = new boolean[N];private static int idx;private static int res;public static void add(int a,int b){e[idx] = b;ne[idx] = h[a];h[a] = idx++;}//能否找到他的妻子public static boolean find(int x){for(int i = h[x]; i != -1; i = ne[i] ){int j = e[i];if(!vis[j]){ // 需要标记一下右边，因为需要考虑当前男生已经选择了这个女生，需要改变别的男生时，别的男生就不能考虑这个女生vis[j] = true;if(match[j] == 0 || find(match[j])){match[j] = x;return true;}}}return false;}public static void main(String[] args) throws IOException {BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));String[] s = reader.readLine().split(" ");n1 = Integer.parseInt(s[0]);n2 = Integer.parseInt(s[1]);m = Integer.parseInt(s[2]);Arrays.fill(h,-1);for(int i = 0; i < m; i++){String[] s1 = reader.readLine().split(" ");int u = Integer.parseInt(s1[0]);int v = Integer.parseInt(s1[1]);add(u,v);}for(int i = 1; i <= n1; i++){Arrays.fill(vis,false);if(find(i)){res++;}}System.out.println(res);}
}
