学步园

思路：简单的二分匹配。

//332K   
0MS
#include <stdio.h>
#include <string.h>
#define N 205

int mat[N][N],vis[N],xM[N],yM[N];
int n,m;

bool DFS(int u)
{
    for (int v =
1;v <= n;v ++)
       
if (mat[u][v] && !vis[v])
       
{
           
vis[v] = 1;
           
if (yM[v] == -1 ||DFS(yM[v]))
           
{
               
yM[v] = u,xM[u] = v;
               
return true;
           
}
       
}
    return
false;
}
int MaxMatch ()
{
    int res =
0;
    memset
(xM,-1,sizeof(xM));
    memset
(yM,-1,sizeof(yM));
    for (int u =
1;u <= n;u ++)
       
if (xM[u] == -1)
       
{
           
memset (vis,0,sizeof(vis));
           
if (DFS(u))
               
res ++;
       
}
    return
res;
}
int main ()
{
    int
u,v,s;
    while
(~scanf ("%d%d",&n,&m))
    {
       
memset (mat,0,sizeof(mat));
       
for (u = 1;u <= n;u ++)
       
{
           
scanf ("%d",&s);
           
while (s --)
           
{