パラレルセット+ベクトルオフセット

http://poj.org/problem?id=1703
Find them, Catch them
Time Limit: 1000MS
 
Memory Limit: 10000K
Total Submissions: 31589
 
Accepted: 9739
Description
The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang Dragon and Gang Snake. However, the police first needs to identify which gang a criminal belongs to. The present question is, given two criminals; do they belong to a same clan? You must give your judgment based on incomplete information. (Since the gangsters are always acting secretly.) 
Assume N (N <= 10^5) criminals are currently in Tadu City, numbered from 1 to N. And of course, at least one of them belongs to Gang Dragon, and the same for Gang Snake. You will be given M (M <= 10^5) messages in sequence, which are in the following two kinds: 
1. D [a] [b] 
where [a] and [b] are the numbers of two criminals, and they belong to different gangs. 
2. A [a] [b] 
where [a] and [b] are the numbers of two criminals. This requires you to decide whether a and b belong to a same gang. 
Input
The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case begins with a line with two integers N and M, followed by M lines each containing one message as described above.
Output
For each message "A [a] [b]"in each case, your program should give the judgment based on the information got before. The answers might be one of "In the same gang.", "In different gangs."and "Not sure yet."
Sample Input

1
5 5
A 1 2
D 1 2
A 1 2
D 2 4
A 1 4

Sample Output

Not sure yet.
In different gangs.
In the same gang.

この町には2人のギャンググループがいて、今N人をあげて、任意の2人に彼らが同じグループでDを入力しているかどうかを聞いています. x yはxを表してyは1つのグループの中でAを入力しません x y xとyが同じグループにいるかどうか、または彼らが同じグループにいるかどうか分からないことを出力します.
図で説明してセットのベクトルオフセットを調べる
1と2を加えると敵、2と3は敵、mark[]で0と1をマークし、以下のように更新します.
a=1;b=2  ==>  x=1,y=2; ==> f[1]=2;
Mark[1]=(mark[1]+mark[2]+1)%2=1;


a=2,b=3 ==> x=2,y=3; ==> f[2]=3;
Mark[2]=(mark[2]+mark[3]+1)%2=1;


今1と3の関係を探しています.
再帰プロセス:
x=1;f[x]=2;x!=f[x]; t=f[x]=2  ====> 
x=2;f[x]=3;x!=f[x], t=f[x]=3  ====>
x=3;f[x]=3; return f[x];  <==f[x]=finde(f[x])==
Mark[2]=(mark[2]+mark[t=3])%2=1; return f[x];  <==f[x]=finde(f[x])==
Mark[1]=(mark[1]+mark[t=2])%2=0; return f[x];
更新完了
発見mark[1]==mark[3];一つです.
 プログラム:

#include"stdio.h"
#include"string.h"
#include"queue"
#include"stack"
#include"math.h"
#include"iostream"
#define M 100005
#define inf 100000000
#define mod 10007
#define eps 1e-10
using namespace std;
int f[M],mark[M];
int finde(int x)
{
    if(x!=f[x])
    {
        int t=f[x];
        f[x]=finde(f[x]);
        mark[x]=(mark[x]+mark[t])%2;
    }
    return f[x];
}
void make(int a,int b)
{
    int x=finde(a);
    int y=finde(b);
    if(x!=y)
    {
        f[x]=y;
        mark[x]=(mark[a]+mark[b]+1)%2;
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m,i;
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
            f[i]=i;
        memset(mark,0,sizeof(mark));
        while(m--)
        {
            int a,b;
            char str[3];
            scanf("%s%d%d",str,&a,&b);
            if(str[0]=='D')
            {
                make(a,b);
            }
            else
            {
                int x=finde(a);
                int y=finde(b);
                if(x!=y)
                    printf("Not sure yet.
");
                else if(mark[a]==mark[b])
                    printf("In the same gang.
");
                else
                    printf("In different gangs.
");
            }
        }
    }
    return 0;
}