HDU 5486 Difference of Clustering 图论

Difference of Clustering

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5486

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

HINT

题意

给你很多个一开始的集合，和结束时候的集合

并且告诉你具体的这些元素是怎么移动的

你要分别算出 分离、合并、1:1这三种操作有多少种

分离就是1个集合变成了多个集合，合并就是多个集合变成了一个集合，1:1就是一个变成了一个

题解：

当成图论做的……

分离操作就是这个集合的边集>1，这个集合连的所有集合的边集都为1

1:1就是这个集合的边集=1，这个集合连的集合的边集也为1

合并操作就是分离操作的逆运算，swap一下，再跑一遍分离就好了

代码：

//qscqesze
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 1000006
#define mod 1000000007
#define eps 1e-9
#define e exp(1.0)
#define PI acos(-1)
const double EP  = 1E-10 ;
int Num;
//const int inf=0x7fffffff;
const ll inf=999999999;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//*************************************************************************************

vector<int> Q;
map<int,int> H;
struct node
{
    int x,y;
};
node p[maxn];
int b[maxn];
vector<int> Q1[maxn];
int main()
{
    int t=read();
    for(int cas=1;cas<=t;cas++)
    {
        int n=read();
        H.clear();
        for(int i=1;i<=n;i++)
        {
            p[i].x=read(),p[i].y=read();
            Q.push_back(p[i].x);
            Q.push_back(p[i].y);
        }

        sort(Q.begin(),Q.end());
        Q.erase(unique(Q.begin(),Q.end()),Q.end());
        int len = Q.size();
        for(int i=0;i<len;i++)
            H[Q[i]]=i;
        for(int i=0;i<len;i++)
            Q1[i].clear(),b[i]=0;
        Q.clear();
        for(int i=1;i<=n;i++)
        {
            Q1[H[p[i].x]].push_back(H[p[i].y]);
        }
        for(int i=0;i<len;i++)
        {
            sort(Q1[i].begin(),Q1[i].end());
            Q1[i].erase(unique(Q1[i].begin(),Q1[i].end()),Q1[i].end());
        }
        for(int i=0;i<len;i++)
        {
            for(int j=0;j<Q1[i].size();j++)
                b[Q1[i][j]]++;
        }
        int ans1 = 0,ans2 = 0,ans3 = 0;
        for(int i=0;i<len;i++)
        {
            if(Q1[i].size()==0)continue;
            if(Q1[i].size()==1)
            {
                if(b[Q1[i][0]]==1)
                    ans3++;
            }
            else
            {
                int flag = 1;
                for(int j=0;j<Q1[i].size();j++)
                {
                    if(b[Q1[i][j]]!=1)
                    {
                        flag = 0;
                        break;
                    }
                }
                if(flag)
                    ans1++;
            }
        }
        for(int i=0;i<len;i++)
            Q1[i].clear(),b[i]=0;
        for(int i=1;i<=n;i++)
        {
            Q1[H[p[i].y]].push_back(H[p[i].x]);
        }
        for(int i=0;i<len;i++)
        {
            sort(Q1[i].begin(),Q1[i].end());
            Q1[i].erase(unique(Q1[i].begin(),Q1[i].end()),Q1[i].end());
        }
        for(int i=0;i<len;i++)
        {
            for(int j=0;j<Q1[i].size();j++)
                b[Q1[i][j]]++;
        }
        for(int i=0;i<len;i++)
        {
            if(Q1[i].size()==0)continue;
            if(Q1[i].size()==1)
            {
                if(b[Q1[i][0]]==1)
                    ans3++;
            }
            else
            {
                int flag = 1;
                for(int j=0;j<Q1[i].size();j++)
                {
                    if(b[Q1[i][j]]!=1)
                    {
                        flag = 0;
                        break;
                    }
                }
                if(flag)
                    ans2++;
            }
        }
        printf("Case #%d: %d %d %d
",cas,ans1,ans2,ans3/2);
    }
}