hdu1083 Courses--最大匹配 & HK算法

原题链接：​​http://acm.hdu.edu.cn/showproblem.php?pid=1083​​

题意：p个课程，n个学生，每门课程都有若干个学生，问是否可以让每门课程都有课代表，可以输出YES。

直接用HK算法就可以，当然这题有点水，用匈牙利也可以。

#define _CRT_SECURE_NO_DEPRECATE 

#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
#define INF 99999999;
using namespace std;

int t;
int dis;
int p, n;
int a[105][305];
int cx[305];
int cy[305];
int dx[305];
int dy[305];
bool vis[305];

bool searchPath()
{
  queue<int> Q;
  memset(dx, -1, sizeof(dx));
  memset(dy, -1, sizeof(dy));
  dis = INF;
  for (int i = 1; i <= p; i++)
  {
    if (cx[i] == -1)
    {
      Q.push(i);
      dx[i] = 0;
    }
  }

  while (!Q.empty())
  {
    int u = Q.front();
    Q.pop();
    if (dx[u] > dis)
      break;
    for (int v = 1; v <= n; v++)
    {
      if (a[u][v] && dy[v] == -1)
      {
        dy[v] = dx[u] + 1;
        if (cy[v] == -1)
          dis = dy[v];
        else
        {
          dx[cy[v]] = dy[v] + 1;
          Q.push(cy[v]);
        }
      }
    }
  }
  return dis != INF;
}

int findPath(int u)
{
  for (int v = 1; v <= n; v++)
  {
    if (!vis[v] && a[u][v] && dy[v] == dx[u] + 1)
    {
      vis[v] = 1;
      if (cy[v] != -1 && dy[v] == dis)
        continue;
      if (cy[v] == -1 || findPath(cy[v]))
      {
        cy[v] = u;
        cx[u] = v;
        return 1;
      }
    }
  }
  return 0;
}

int maxMatch()
{
  int ans = 0;
  memset(cx, -1, sizeof(cx));
  memset(cy, -1, sizeof(cy));
  while (searchPath())
  {
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= p; i++)
    {
      if (cx[i] == -1)
        ans += findPath(i);
    }
  }
  return ans;
}

int main()
{
  scanf("%d", &t);
  while (t--)
  {
    scanf("%d%d", &p, &n);
    memset(a, 0, sizeof(a));
    
    for (int i = 1; i <= p; i++)
    {
      int num, x;
      scanf("%d", &num);
      for (int j = 1; j <= num; j++)
      {
        scanf("%d", &x);
        a[i][x] = 1;
      }
    }
    printf("%s\n", maxMatch() == p ? "YES" : "NO");
  }

  return 0;
}