二分匹配

#include <stdio.h>
#include <string.h>

const int MAXN=510;
int uN,vN;//u,v数目
int g[MAXN][MAXN];
int yM[MAXN];
int m[MAXN][MAXN];
bool used[MAXN];
int xM[MAXN];

bool dfs(int u)//从左边开始找增广路径
{
    int v;
    for(v = 0; v < vN; v++)//这个顶点编号从0开始，若要从1开始需要修改
      if(g[u][v] && !used[v])
      {
          used[v] = true;
          if(yM[v] == -1 || dfs(yM[v]))
          {//找增广路，反向
              yM[v] = u;
			  xM[u] = v;
              return true;
          }
      }
    return false;
}
int hungary()
{
    int res = 0;
    int u;
    memset(yM ,-1, sizeof(yM));
	memset(xM ,-1, sizeof(xM));
    for(u = 0; u < uN; u++)
    {
        memset(used, 0, sizeof(used));
        if (dfs(u)) res++;
    }
    return res;
}
int main()
{
	int t1, t2, t3;

	while (scanf("%d%d", &uN, &vN) != EOF)
	{
		sum = 0;
		memset(g, 0, sizeof(g));
		memset(m, 0, sizeof(m));
		for (int i = 0; i < uN; i++)
			for (int j = 0; j < vN; j++)
			{
				scanf("%d", &t1);
				m[i][j] = t1;
				if (t1 != 0)
					g[i][j] = 1;
			}
		
		printf("%d\n", hungary());
		for (int i = 0; i < uN; i++)
        {
			printf ("%d ", xM[i]);
        }
		printf("\n");
	}
	return 0;
}