lightoj 1009 Back to Underworld （种类并查集）

The Vampires and Lykans are fighting each other to death. The war has become so fierce that, none knows who will win. The humans want to know who will survive finally. But humans are afraid of going to the battlefield.

So, they made a plan. They collected the information from the newspapers of Vampires and Lykans. They found the information about all the dual fights. Dual fight means a fight between a Lykan and a Vampire. They know the name of the dual fighters, but don't know which one of them is a Vampire or a Lykan.

So, the humans listed all the rivals. They want to find the maximum possible number of Vampires or Lykans.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case contains an integer n (1 ≤ n ≤ 10⁵), denoting the number of dual fights. Each of the nextn lines will contain two different integers u v (1 ≤ u, v ≤ 20000) denoting there was a fight betweenu and v. No rival will be reported more than once.

Output

For each case, print the case number and the maximum possible members of any race.

Sample Input

2

2

1 2

2 3

3

1 2

2 3

4 2

Sample Output

Case 1: 2

Case 2: 3

Hint

Dataset is huge, use faster I/O methods.

种类并查集。

group[i]代表i和他的祖先的关系，group[i]=1代表是同类，=0代表不是同类。

#include<bits/stdc++.h>#define N 20000+10using namespace std;int f[N],group[N];int in[N],num[N],flag[N];int Find(int x){    if(x==f[x])        return x;    int tx=f[x];    f[x]=Find(f[x]);    group[x]=(group[x]+group[tx]+1)%2;    return f[x];}int main(){    int t,m,n,a,b;    int cas=1;    scanf("%d",&t);    while(t--)    {        scanf("%d",&m);        for(int i=1; i<N; i++)        {            f[i]=i;            group[i]=1;            in[i]=0;            flag[i]=0;            num[i]=1;        }        n=-1;        for(int i=1; i<=m; i++)        {            scanf("%d%d",&a,&b);            n=max(a,n);            n=max(b,n);            in[a]=1,in[b]=1;            int ta=Find(a);            int tb=Find(b);            if(ta!=tb)            {                f[ta]=tb;                num[tb]+=num[ta];//集合里的元素整体要归到祖先下面                group[ta]=(group[b]-group[a]+2)%2;            }        }        for(int i=1;i<=n;i++)        {            if(in[i])            {                int ti=Find(i);                if(group[i]==0)                    flag[ti]++;            }        }        int ans=0;        for(int i=1;i<=n;i++)//计算每一个祖先下面的儿子们与祖先的关系，累加        {            if(in[i]&&f[i]==i)            ans+=max(flag[i],num[i]-flag[i]);        }        printf("Case %d: %d\n",cas++,ans);    }    return 0;}