hdu 5486 Difference of Clustering(合肥网赛)

Difference of Clustering

                                                                                      Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                                             Total Submission(s): 74    Accepted Submission(s): 17

Problem Description

Given two clustering algorithms, the old and the new, you want to find the difference between their results. 
A clustering algorithm takes many  member entities  as input and partition them into  clusters . In this problem, a member entity must be clustered into exactly one cluster. However, we don’t have any pre-knowledge of the clusters, so different algorithms may produce different number of clusters as well as different cluster IDs. One thing we are sure about is that the  memberIDs  are stable, which means that the same member ID across different algorithms indicates the same member entity.
To compare two clustering algorithms, we care about three kinds of relationship between the old clusters and the new clusters: split, merge and 1:1. Please refer to the figure below.

Let’s explain them with examples. Say in the old result, m0, m1, m2 are clustered into one cluster c0, but in the new result, m0 and m1 are clustered into c0, but m2 alone is clustered into c1. We denote the relationship like the following:
●  In the old, c0 = [m0, m1, m2]
●  In the new, c0 = [m0, m1], c1 = [m2]
There is no other members in the new c0 and c1. Then we say the old c0 is split into new c0 and new c1. A few more examples:
●  In the old, c0 = [m0, m1, m2]
●  In the new, c0 = [m0, m1, m2]. 
This is 1:1.
●  In the old, c0 = [m0, m1], c1 = [m2]
●  In the new, c0 = [m0, m1, m2]
This is merge. Please note, besides these relationship, there is another kind called “n:n”:
●  In the old, c0 = [m0, m1], c1 = [m2, m3]
●  In the new, c0 = [m0, m1, m2], c1 = [m3]
We don’t care about n:n. 
In this problem, we will give you two sets of clustering results, each describing the old and the new. We want to know the total number of splits, merges, and 1:1 respectively.

 

Input

The first line of input contains a number  T  indicating the number of test cases ( T≤100 ).
Each test case starts with a line containing an integer  N  indicating the number of member entities ( 0≤N≤106  ). In the following  N  lines, the  i -th line contains two integers c1 and c2, which means that the member entity with ID i is partitioned into cluster c1 and cluster c2 by the old algorithm and the new algorithm respectively. The cluster IDs c1 and c2 can always fit into a 32-bit signed integer.

 

Output

For each test case, output a single line consisting of “Case #X: A B C”.  X  is the test case number starting from 1.  A ,  B , and  C  are the numbers of splits, merges, and 1:1s.

 

Sample Input

   
   
   
   

    
    
    
    2
3
0 0
0 0
0 1
4
0 0
0 0
1 1
1 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 1 0 0
Case #2: 0 0 2
   
   
   
   

 

  解题思路：

       首先要离散化，然后将每对old和new建立一条边，重复的边要去掉，并统计每个点的度，于是就会发现1:1的

 肯定是这两个顶点度有一条边，并且两个顶点的度都为1，1:n的情况，即split，odd的顶点的度肯定大于1，并且

  与之相连的顶点的度都为1，于是solit和1:1的就解决了，再反过来，以new的顶点为边的起点，就可以解决merge

 

  的情况。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;


struct EDGE
{
    int to,next;
    EDGE(int a,int b)
    {
        to=a;
        next=b;
    }
};

vector<EDGE> edge;
pair<int,int> node[1000010];
int du[2000010];
int n,ne=0;
int num[1000010],num2[1000010];
int head[2000010];


void addedge(int u,int v)
{
    //cout<<u<<" "<<v<<endl;
    edge.push_back(EDGE(v,head[u]));
  //  edge[ne].to=v;
//    edge[ne].next=head[u];
    head[u]=ne++;
}

bool cmp(pair<int,int> a,pair<int,int> b)
{
    if(a.first==b.first) return a.second<b.second;
    return a.first<b.first;
}

int uni()
{
    int len=0;
    sort(node,node+n,cmp);
    for(int i=0;i<n;i++)
    {
        if(i==n-1||node[i].first!=node[i+1].first||node[i].second!=node[i+1].second)
            node[len++]=node[i];
    }
    return len;
}


int main()
{
  //  freopen("D:\\in.txt","r",stdin);
    int T,i,j,cnt=1;
    cin>>T;
    while(T--)
    {
        memset(du,0,sizeof(du));
        ne=0;
        edge.clear();
        memset(head,-1,sizeof(head));
        cin>>n;
        for(i=0;i<n;i++)
        {
            scanf("%d %d",&node[i].first,&node[i].second);
            num[i]=node[i].first;
            num2[i]=node[i].second;
        }
        sort(num,num+n);
        sort(num2,num2+n);
        int len=unique(num,num+n)-num;
        int len2=unique(num2,num2+n)-num2;
        map<int,int> lst,lst2;
        for(i=1;i<=len;i++) lst[num[i-1]]=i;
        for(i=1;i<=len2;i++) lst2[num2[i-1]]=i+len;
        int len3=uni();
        for(i=0;i<len3;i++)
        {
            int x=lst[node[i].first],y=lst2[node[i].second];
       //     cout<<x<<" "<<y<<endl;
            du[x]++,du[y]++;
            addedge(x,y);
            addedge(y,x);
        }
        //cout<<"ok\n";
        int ans1=0,ans2=0,ans3=0;
        for(i=1;i<=len;i++)
        {
            if(du[i]==1)
            {
                int to=edge[head[i]].to;
        //        cout<<to<<endl;
                if(du[to]==1) ans3++;
            }
            else if(du[i]>1)
            {
                int flag=1;
                for(int t=head[i];t!=-1;t=edge[t].next)
                {
                    if(du[edge[t].to]>1)
                    {
                        flag=0;
                        break;
                    }
                }
         //       cout<<flag<<" "<<i<<endl;
                if(flag) ans1++;
            }
        }
        for(i=len+1;i<=len+len2;i++)
        {
            if(du[i]>1)
            {
                int flag=1;
                for(int t=head[i];t!=-1;t=edge[t].next)
                {
                    if(du[edge[t].to]>1)
                    {
                        flag=0;
                        break;
                    }
                }
                if(flag) ans2++;
            }
        }
        printf("Case #%d: %d %d %d\n",cnt++,ans1,ans2,ans3);
    }
}