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,0,sizeof(q));
    int start=0,end=1;
    q[start]=1;
    memset(judge,-1,sizeof(judge));
    judge[1]=0;
    while (start<end)
    {
        int w=q[start];
        for (int i=0;i<line[w].size();i++)
        {
            int e=line[w][i];
            if (judge[e]==-1){
                    judge[e]=(judge[w]+1)%2;
                    q[end++]=e;
            }
            else{
                if (judge[e]==judge[w]) return 0;
            }
        }
        start++;
    }
    return 1;
}
int sreach(int x)
{
    int i,j;
    for (j=1;j<=n;j++){
    for (i=0;i<line[j].size();i++)
    {
        if (line[j][i]==x&&!used[j])
        {
            used[j]=1;
            if (!p[j]||sreach(p[j]))
            {
                p[j]=x;
                return 1;
            }
        }
    }
    }
    return 0;
}
int main()
{
    int t,i,m,a,b;
    while (~scanf("%d %d",&n,&m))
    {
        for (i=0;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,0,sizeof(p));
        if (!find()){
            printf("No
");continue;
        }
        int ans=0;
        for (i=1;i<=n;i++)
        {
            memset(used,0,sizeof(used));
            if (sreach(i)) ans++;
        }
        printf("%d
",ans/2);
    }
    return 0;
}

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=0;i<=200;i++){
        father[i]=i;rank[i]=0;
    }
}
int find(int x)
{
    if (x==father[x]) return father[x];
    int t=find(father[x]);
    rank[x]=(rank[x]+rank[father[x]])%2;
    father[x]=t;
    return father[x];
}
int sreach(int x,int n)
{
    int i;
    for (i=1;i<=n;i++)
    {
        if (line[i][x]&&!used[i])
        {
           used[i]=1;
           if (!p[i]||sreach(p[i],n))
           {
               p[i]=x;
               return 1;
           }
        }
    }
    return 0;
}
int main()
{
    int n,m,a,b,i;
    while (~scanf("%d %d",&n,&m))
    {
        give();
        int flag=0;
        memset(p,0,sizeof(p));
        memset(line,0,sizeof(line));
        while (m--)
        {
            scanf("%d %d",&a,&b);
            line[a][b]=line[b][a]=1;
            int sx=find(a);
            int sy=find(b);
            if (sx!=sy) {
                rank[sx]=(rank[a]+rank[b]+1)%2;
                father[sx]=sy;
            }
            else {
                if (rank[a]==rank[b]) flag=1;
            }
        }
        if (flag==1){
            printf("No
");continue;
        }
        int ans=0;
        for (i=1;i<=n;i++)
        {
            memset(used,0,sizeof(used));
            if (sreach(i,n)) ans++;
        }
        printf("%d
",ans/2);
    }
    return 0;
}