poj 1703（带权并查集）

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.

Source

POJ Monthly--2004.07.18

这题自己当时想的时候总是想方设法去区分每个点是属于哪个集团，在第一次遇到D时就随机分两个集团，可是发现这样根本无法继续下去，因为如果出现的两个数字以前都没有出现过的话，怎么也无法确定各自集团的，后来网上看到了很多方法，但我觉得这个是最好的，最简单的思想，保留所有的可能性即可，没必要去纠结谁属于谁，把所有的可能都存储下来，每个点存储a,和a+n,a代表其属于某个集团，a+n属于另外一个集团，按照并查集存储并执行就好了。

 1 #include<iostream>
 2 #include<cstdio>
 3
 4 using namespace std;
 5
 6 int parent[200005];
 7
 8 int Getparent(int x)
 9 {
10     if(x!=parent[x])
11         parent[x] = Getparent(parent[x]);
12     return parent[x];
13 }
14
15 void Union(int x,int y)
16 {
17     int a = Getparent(x),b = Getparent(y);
18     if(a!=b)
19         parent[b] = a;
20 }
21
22 int main()
23 {
24     int t;
25     while(scanf("%d",&t)!=EOF)
26     {
27         while(t--)
28         {
29             int n,m;
30             scanf("%d%d",&n,&m);
31             for(int i=0;i<n+n+1;i++)
32             {
33                 parent[i] = i;
34             }
35             for(int i=0;i<m;i++)
36             {
37                 char s;
38                 int a,b;
39                 getchar();
40                 scanf("%c %d %d",&s,&a,&b);
41                 if(s==‘A‘)
42                 {
43                     if(Getparent(a) == Getparent(b)||Getparent(a+n)==Getparent(b+n))
44                         printf("In the same gang.\n");
45                     else if(Getparent(a) == Getparent(b+n)||Getparent(b) == Getparent(a+n))
46                         printf("In different gangs.\n");
47                     else
48                         printf("Not sure yet.\n");
49                 }
50                 else
51                 {
52                     Union(a,b+n);
53                     Union(a+n,b);
54                 }
55             }
56         }
57     }
58     return 0;
59 }