POJ 1703 Find them, Catch them(种类并查集）

Find them, Catch them
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 44682 Accepted: 13757
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.

题意：城市里有两个黑帮团伙，现在给出N个人，问任意两个人他们是否在同一个团伙
输入D x y代表x于y不在一个团伙里
输入A x y要输出x与y是否在同一团伙或者不确定他们在同一个团伙里

题解：种类并查集只是带权并查集再弄个%取余操作而已，然后余数就表示他属于哪个种类。
这题只有两个种类，也就是只有0和1两种， 对于两个不同的种类，那么之间的权值是相差1的，所以按照带权并查集的方法做加上1，然后取余2即可。

代码：

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <string>#include <vector>using namespace std;const int maxn = 1e5 + 5;int f[maxn];int relation[maxn];int m,n;void init(int){    for(int i=0;i<n;i++)    {        f[i] = i;        relation[i] = 0;    }}int find(int x){    if(x == f[x])        return x;    int tmp = find(f[x]);    relation[x] = (relation[x] + relation[f[x]])%2;//根据儿子与父亲的关系和父亲与爷爷的关系推出儿子与爷爷的关系    f[x] = tmp;    return f[x];}void mix(int x,int y){    int fx = find(x);    int fy = find(y);    if(fx == fy)        return;        f[fx] = fy;        relation[fx] = (relation[y] + relation[x] +1)%2;        return;}int main(){    int t;    cin>>t;    while(t--)    {        cin>>n>>m;        init(n);        char c;        int x,y;        getchar();        for(int i=1;i<=m;i++)        {            scanf("%c%d%d%*c", &c, &x, &y);            if(c == 'A'&&find(x) == find(y))            {                if(relation[x] == relation[y])                    cout << "In the same gang." << endl;                else cout << "In different gangs." << endl;            }            else  if(c == 'D')            {                mix(x,y);            }            else cout<<"Not sure yet."<<endl;        }    }    return 0;}