hdu 2444(二分图) The Accomodation of Students

http://acm.hdu.edu.cn/showproblem.php?pid=2444

大意是给定n个学生，他们之间可能互相认识，首先判断能不能将这些学生分为两组，使组内学生不认识；

现想将学生两两分组，且保证每一组的学生都认识，这样分组可达到的最大组数为多大？

判断二分图,然后求匈牙利算法求最大匹配数

染色法判断二分图(脑抽用vector跑的特别慢)

 #include<cstdio>
 #include<cstring>
 #include<vector>
 using namespace std;
 #define M 300
 vector<int>line[M];
 int judge[M],p[M],q[M],used[M];
 int n;
 int find()
 {
     memset(q,,sizeof(q));
     int start=,end=;
     q[start]=;
     memset(judge,-,sizeof(judge));
     judge[]=;
     while (start<end)
     {
         int w=q[start];
         for (int i=;i<line[w].size();i++)
         {
             int e=line[w][i];
             if (judge[e]==-){
                     judge[e]=(judge[w]+)%;
                     q[end++]=e;
             }
             else{
                 if (judge[e]==judge[w]) return ;
             }
         }
         start++;
     }
     return ;
 }
 int sreach(int x)
 {
     int i,j;
     for (j=;j<=n;j++){
     for (i=;i<line[j].size();i++)
     {
         if (line[j][i]==x&&!used[j])
         {
             used[j]=;
             if (!p[j]||sreach(p[j]))
             {
                 p[j]=x;
                 return ;
             }
         }
     }
     }
     return ;
 }
 int main()
 {
     int t,i,m,a,b;
     while (~scanf("%d %d",&n,&m))
     {
         for (i=;i<=n;i++)
             line[i].clear();
         while (m--)
         {
             scanf("%d %d",&a,&b);
             line[a].push_back(b);
             line[b].push_back(a);
         }
         memset(p,,sizeof(p));
         if (!find()){
             printf("No\n");continue;
         }
         int ans=;
         for (i=;i<=n;i++)
         {
             memset(used,,sizeof(used));
             if (sreach(i)) ans++;
         }
         printf("%d\n",ans/);
     }
     return ;
 }

关系并查集判断二分图,类似于hdu 1829

#include<cstdio>
#include<cstring>
using namespace std;
#define M 300
int father[M],line[M][M],used[M],p[M];
int n,rank[M];
void give()
{
    for (int i=;i<=;i++){
        father[i]=i;rank[i]=;
    }
}
int find(int x)
{
    if (x==father[x]) return father[x];
    int t=find(father[x]);
    rank[x]=(rank[x]+rank[father[x]])%;
    father[x]=t;
    return father[x];
}
int sreach(int x,int n)
{
    int i;
    for (i=;i<=n;i++)
    {
        if (line[i][x]&&!used[i])
        {
           used[i]=;
           if (!p[i]||sreach(p[i],n))
           {
               p[i]=x;
               return ;
           }
        }
    }
    return ;
}
int main()
{
    int n,m,a,b,i;
    while (~scanf("%d %d",&n,&m))
    {
        give();
        int flag=;
        memset(p,,sizeof(p));
        memset(line,,sizeof(line));
        while (m--)
        {
            scanf("%d %d",&a,&b);
            line[a][b]=line[b][a]=;
            int sx=find(a);
            int sy=find(b);
            if (sx!=sy) {
                rank[sx]=(rank[a]+rank[b]+)%;
                father[sx]=sy;
            }
            else {
                if (rank[a]==rank[b]) flag=;
            }
        }
        if (flag==){
            printf("No\n");continue;
        }
        int ans=;
        for (i=;i<=n;i++)
        {
            memset(used,,sizeof(used));
            if (sreach(i,n)) ans++;
        }
        printf("%d\n",ans/);
    }
    return ;
}