POJ 1703-Find them, Catch them（并查集）

绝地灬酷狼 2022-06-16 03:22 84阅读 0赞

Find them, Catch them

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;45008</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;13862</td> 
  </tr> 
 </tbody> 
</table>

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.

Source

[POJ Monthly--2004.07.18][]

## 题目意思： ##

一共有俩犯罪团伙，N个人中有人可能是罪犯，D a b表示a和b属于同一个犯罪团伙，A a b表示询问a和b的团伙关系。

## 解题思路： ##

反正就俩犯罪团伙，标记set\[a\]和set\[a+n\]分别是两个犯罪团伙。

合并时如果不是其中一派，那就是另外一派；查询时如果不是其中一派还可能自成一派。

#include <iostream>
    #include <cstring>
    #include <cstdio>
    using namespace std;
    const long N=100000+5;
    int m,n;
    int set[N+N];
    int set_find(int d)
    {
        if(set[d]<0) return d;
        return set[d]=set_find(set[d]);
    }
    void unite(int p,int q)
    {
        p=set_find(p);
        q=set_find(q);
        if(p!=q) set[p]=q;
    }
    int main()
    {
        int T,i;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d%d",&n,&m);
            memset(set,-1,sizeof(set));//初始化时各成一派
            for(i=0; i<m; ++i)
            {
                int a,b;
                char s[5];
                scanf("%s%d%d",s,&a,&b);
                if(s[0]=='A')//询问
                {
                    if(set_find(a)!=set_find(b)&&set_find(a)!=set_find(b+n))
                        printf("%s\n","Not sure yet.");
                    else if(set_find(a)==set_find(b))
                        printf("%s\n","In the same gang.");
                    else
                        printf("%s\n","In different gangs.");
                }
                else//合并
                {
                    unite(a,b+n);
                    unite(b,a+n);
                }
            }
        }
        return 0;
    }

[POJ Monthly--2004.07.18]: http://poj.org/searchproblem?field=source&key=POJ+Monthly--2004.07.18