HDU 5489 Difference of Clustering 图论

Difference of Clustering

[align=left]Problem Description[/align]
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.

///1085422276
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<bitset>
#include<set>
#include<vector>
using namespace std ;
typedef __int64 ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define memfy(a)  memset(a,-1,sizeof(a));
#define TS printf("111111\n");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define mod 1000000007
#define inf 100000000
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;
}
//****************************************

#define maxn 1000000+6
struct ss
{
int to,next;
} e[maxn];
struct node
{
int x,index;//0,1;
};
int head[maxn],n,a,b,t,in[maxn][2],A,B,C;
map<pair<int ,int >,int  >mp;
map<int ,int >vis,vis2;
map<int ,vector<int > >mpp,mpp2;
vector<int >V1,V2;
vector<int >::iterator it;;
int main()
{

int T=read();
int oo=1;
while(T--)
{
// init();
scanf("%d",&n);
mp.clear();
V1.clear();
V2.clear();
mpp2.clear();
mpp.clear();
vis.clear();
vis2.clear();
int k=0;
FOR(i,1,n)
{
scanf("%d%d",&a,&b);
if(mp[make_pair(a,b)])continue;
mpp[a].push_back(b);
mpp2[b].push_back(a);
if(!vis[a])
V1.push_back(a);
if(!vis2[b])
V2.push_back(b);
vis[a]=1;
vis2[b]=1;
mp[make_pair(a,b)]=1;
}
A=0;
B=0;
C=0;
int sum;
for(int i=0; i<V1.size(); i++)
{
sum=0;
for(it=mpp[V1[i]].begin(); it!=mpp[V1[i]].end(); it++)
{
sum+=mpp2[*it].size();
}
if(sum==mpp[V1[i]].size())
{
if(sum==1)
C++;
else
{
A++;
}
}
}
for(int i=0; i<V2.size(); i++)
{
sum=0;
for(it=mpp2[V2[i]].begin(); it!=mpp2[V2[i]].end(); it++)
{
sum+=mpp[*it].size();
}
if(sum==mpp2[V2[i]].size())
{  //cout<<mpp[V2[i]].size()<<endl;
if(sum==1)
C++;
else
{
B++;
}
}
}
printf("Case #%d: ",oo++);
cout<<A<<" "<<B<<" "<<C/2<<endl;

}
return 0;
}

代码