POJ 1703 Find them, Catch them

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24844		Accepted: 7457

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.

简单的并查集问题，有一组测试数据需要注意
1
2 1
A 1 2
应该输出"In different gangs."，因为两个帮派至少都有一人。

 1 #include<iostream>
 2 #include<cstdio>
 3 
 4 using namespace std;
 5 
 6 typedef struct node
 7 {
 8     long father;
 9     long relation;  //相同为0，不同为1
10 } NODE;
11 
12 NODE criminal[100050];
13 
14 void MakeSet(long x)
15 {
16     for(int i=1;i<=x;i++)
17     {
18         criminal[i].father=i;
19         criminal[i].relation=0;
20     }
21 }
22 
23 long Find(long x)
24 {
25     if(criminal[x].father==x)
26         return x;
27     long pre_father=criminal[x].father;
28     criminal[x].father=Find(criminal[x].father);
29     criminal[x].relation=(criminal[x].relation+criminal[pre_father].relation)%2;
30     return criminal[x].father;
31 }
32 
33 void Merge(long x,long y,int r)
34 {
35     criminal[x].father=y;
36     criminal[x].relation=r;
37 }
38 
39 int main()
40 {
41     int t;
42     long m,n;
43 
44     scanf("%d",&t);
45 
46     while(t--)
47     {
48         char c;
49         long a,b,af,bf;
50         scanf("%ld%ld",&n,&m);
51         MakeSet(n);
52         for(long i=1;i<=m;i++)
53         {
54             getchar();
55             scanf("%c %ld %ld",&c,&a,&b);
56             af=Find(a);
57             bf=Find(b);
58             if(c=='A')
59             {
60                 if(n==2)
61                     printf("In different gangs.\n");
62                 else
63                 {
64                     if(af==bf)
65                     {
66                         if(criminal[a].relation==criminal[b].relation)
67                             printf("In the same gang.\n");
68                         else
69                             printf("In different gangs.\n");
70                     }
71                     else
72                         printf("Not sure yet.\n");
73                 }
74             }
75             else
76             {
77                 if(af!=bf)
78                     Merge(af,bf,(criminal[a].relation+criminal[b].relation+1)%2);
79             }
80         }
81     }
82 
83     return 0;
84 }

[C++]

 

转载于:https://www.cnblogs.com/lzj-0218/p/3181298.html