Find them, Catch them
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
题目类型：带权并查集
算法分析：简化的带权并查集，rel数组中0表示两个人在同一个帮派中，1表示两个人不在同一个帮派中。注意推导出的求两个根节点关系的公式
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#include <iostream> #include <fstream> #include <algorithm> #include <iomanip> #include <cstring> #include <cstdio> #include <cmath> #include <map> #include <string> #include <vector> #include <stack> #include <queue> #include <set> #include <list> #include <ctime> using namespace std; const int maxn = 100000 + 66; int parent[maxn], rel[maxn]; int UnFind (int val) { if (parent[val] != val) { int temp = parent[val]; parent[val] = UnFind (parent[val]); rel[val] = (rel[val] + rel[temp]) % 2; } return parent[val]; } int main() { // freopen ("aaa.txt", "r", stdin); int t; scanf ("%d", &t); while (t--) { int n, m; scanf ("%d%d", &n, &m); int i; for (i = 0; i < n + 6; i++) { parent[i] = i; rel[i] = 0; } for (i = 0; i < m; i++) { char flag[3]; int a, b; scanf ("%s%d%d", &flag, &a, &b); int p_a = UnFind (a); int p_b = UnFind (b); if (flag[0] == 'A') { if (p_a != p_b) printf ("Not sure yet.\n"); else { if (rel[a] != rel[b]) printf ("In different gangs.\n"); else printf ("In the same gang.\n"); } } else { parent[p_b] = p_a; rel[p_b] = (rel[a] - rel[b] + 1) % 2; } } } return 0; } |
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