HDU 5486(Difference of Clustering-聚类)

Difference of Clustering

给定10^6个关系

现在要求从左到右1对多，多对1，1:1的情况的数量

但是n:n的情况要排除

直接离散，求连通块

#include<bits/stdc++.h>
using namespace std;
#define
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typedef long long ll;

int u[MAXN],v[MAXN],n,a[MAXN];
map<int,int> h;

int sz;
bool b[MAXN];
int cal1,cal2;

class link_table
{
public: 
    void mem()
    {
        MEM(pre) MEM(edge) MEM(pre) size=1;
    }
    int edge[MAXM],next[MAXM],pre[MAXN],size;

    void addedge(int u,int v)
    {
        edge[++size]=v;
        next[size]=pre[u];  
        pre[u]=size;
    }
    void addedge2(int u,int v){addedge(u,v);addedge(v,u);}


    void dfs(int x)
    {
        b[x]=1;
        if (x<=n) ++cal1;else ++cal2;
        Forp(x) {
            int v=edge[p];
            if (!b[v]) dfs(v);
        }

    } 
}St;
int main()
{
    freopen("C.in","r",stdin);
//  freopen(".out","w",stdout);

    int T; cin>>T;
    For(kcase,T) {
        int A=0,B=0,C=0;
        cin>>n;
        For(i,n) {
            int x,y;
            scanf("%d%d",&x,&y);    
            u[i]=x,v[i]=y;
        }
        sz=0;
        For(i,n) a[++sz]=u[i],a[++sz]=v[i];
        sort(a+1,a+1+sz); 
        sz=unique(a+1,a+1+sz)-(a+1);
        h.clear();
        For(i,sz) h[a[i]]=i;
        For(i,n) u[i]=h[u[i]],v[i]=h[v[i]];

        St.mem();
        For(i,n)
        {
            St.addedge2(u[i],v[i]+n);
        }
        For(i,n*2) b[i]=0;
        For(i,n*2) {
            if (!b[i]) 
            {
                cal1=cal2=0;
                St.dfs(i);
                if (cal1==1&&cal2==1) C++;
                else if (cal1==1&&cal2>1) A++;
                else if (cal2==1&&cal1>1) B++;

            }
        }       
        printf("Case #%d: %d %d %d\n",kcase,A,B,C);
    }


    return 0;
}