HDU 3829 Cat VS Dog（二分图，4级）

最新推荐文章于 2022-03-15 14:14:57 发布

剑不飞

最新推荐文章于 2022-03-15 14:14:57 发布

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C - Cat VS Dog

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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Appoint description:

Description

The zoo have N cats and M dogs, today there are P children visiting the zoo, each child has a like-animal and a dislike-animal, if the child's like-animal is a cat, then his/hers dislike-animal must be a dog, and vice versa.
Now the zoo administrator is removing some animals, if one child's like-animal is not removed and his/hers dislike-animal is removed, he/she will be happy. So the administrator wants to know which animals he should remove to make maximum number of happy children.

Input

The input file contains multiple test cases, for each case, the first line contains three integers N <= 100, M <= 100 and P <= 500.
Next P lines, each line contains a child's like-animal and dislike-animal, C for cat and D for dog. (See sample for details)

Output

For each case, output a single integer: the maximum number of happy children.

Sample Input

Sample Output

1
3

Hint

Case 2: Remove D1 and D2, that makes child 1, 2, 3 happy.

思路：纯二分图，根据人来建图，凡是互斥的建边，建双向方便，但结果需除2

#include<cstdio>
#include<cstring>
#include<iostream>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mm=1550;
class ppp
{
public:
  int dog,cat;
  bool flag;
}f[mm];
int N,M,P;
char s[6];
int head[mm],edge;
class Edge
{
  public:int v,next;
}e[mm*mm*2];
void init()
{
  edge=0;clr(head,-1);
}
void add(int u,int v)
{
  e[edge].v=v;e[edge].next=head[u];head[u]=edge++;
}
int X[mm];bool T[mm];
bool dfs(int u)
{ int v;
  for(int i=head[u];~i;i=e[i].next)
  {
    v=e[i].v;
    if(T[v])continue;
    T[v]=1;
    if(X[v]==-1||dfs(X[v]))
    {
      X[v]=u;return 1;
    }
  }
  return 0;
}
void getans()
{ int ret=0;
  FOR(i,0,P)X[i]=-1;
  FOR(i,1,P)
  {
    clr(T,0);
    if(dfs(i))
      ++ret;
  }
  printf("%d\n",P-ret/2);
}
int get(char*s)
{
  int z=0;
  for(int i=1;s[i];++i)
    z=z*10+s[i]-'0';
    return z;
}
bool judge(int x,int y)
{
  bool flag=f[x].flag^f[y].flag;
  if(!flag)return 0;
  if(f[x].dog==f[y].dog||f[x].cat==f[y].cat)return 1;
  return 0;
}
int main()
{
  while(~scanf("%d%d%d",&N,&M,&P))
  { init();
    FOR(i,1,P)
    { bool dog;
      int z;
      scanf("%s",s);
      z=get(s);
      if(s[0]=='D')dog=1;
      else dog=0;

      f[i].flag=dog;
      if(dog)f[i].dog=z;
      else f[i].cat=z;
      scanf("%s",s);
      if(dog)f[i].cat=get(s);
      else f[i].dog=get(s);
    }
    FOR(i,1,P)FOR(j,i+1,P)
    if(i!=j&&judge(i,j))
      {add(i,j),add(j,i);
//       cout<<i<<" "<<j<<endl;
      }
    getans();
  }
}