hdu2444 The Accomodation of Students--二分图判断 & 最大匹配数

原题链接：​​http://acm.hdu.edu.cn/showproblem.php?pid=2444​​

题意：给定n个人的m对关系，判断是否可以组成二分图，不可以的话输出No；不然输出最大匹配数。

#define _CRT_SECURE_NO_DEPRECATE 

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

int n, m;
int a[205][205];
int color[205];
int match[205];
int vis[205];

bool bfs(int s)
{
  queue<int> Q;
  Q.push(s);
  color[s] = 1;

  while (!Q.empty())
  {
    int u = Q.front();
    Q.pop();
    for (int v = 1; v <= n; v++)
    {
      if (a[u][v] && color[v] == -1)
      {
        color[v] = 1 - color[u];
        Q.push(v);
      }
      else if (a[u][v] && color[v] == color[u])
        return false;
    }
  }
  return true;
}

/* 要从1到n一个一个判断，因为可能图是不连通的，也就是起点就是一个独立点 */
bool judge()
{
  memset(color, -1, sizeof(color));
  for (int i = 1; i <= n; i++)
    if (color[i] == -1 && !bfs(i))
      return false;
  return true;
}

int dfs(int u)
{
  for (int v = 1; v <= n; v++)
  {
    if (a[u][v] && !vis[v])
    {
      vis[v] = 1;
      if (match[v] == -1 || dfs(match[v]))
      {
        match[v] = u;
        match[u] = v;
        return 1;
      }
    }
  }
  return 0;
}

int hungarian()
{
  int ans = 0;
  memset(match, -1, sizeof(match));
  for (int i = 1; i <= n; i++)
  {
    if (match[i] == -1)
    {
      memset(vis, 0, sizeof(vis));
      ans += dfs(i);
    }
  }
  return ans;
}

int main()
{
  int x, y;
  while (~scanf("%d%d", &n, &m))
  {
    memset(a, 0, sizeof(a));

    for (int i = 1; i <= m; i++)
    {
      scanf("%d%d", &x, &y);
      a[x][y] = a[y][x] = 1;
    }
  
    if (!judge())
      printf("No\n");
    else
      printf("%d\n", hungarian());
  }
  return 0;
}