POJ 1703 Find them, Catch them

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.

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标算：带偏移量的并查集。
然而我并不会（wa飞了）。
民间做法：好像叫影子集（雾
就是说，建两个并查集，一个放同伙，一个放敌人。
于是就是简单的并查集了。（妙啊）

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#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int maxn=100005;int t,n,m,f[2*maxn];int fnd(int x){    if(f[x]!=x)        f[x]=fnd(f[x]);    return f[x];}int main(){    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=1;i<=2*n;i++)            f[i]=i;        int x,y;        char ch[11];        while(m--)        {            scanf("%s%d%d",ch,&x,&y);            if(ch[0]=='D')            {                f[fnd(x+n)]=fnd(y);                f[fnd(y+n)]=fnd(x);            }            else            {                if(fnd(x)==fnd(y))                    printf("In the same gang.\n");                else if(fnd(x+n)==fnd(y)||fnd(y+n)==fnd(x))                    printf("In different gangs.\n");                else                    printf("Not sure yet.\n");            }        }    }    return 0;}