Find them, Catch them（种类并查集）

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. 思路：这道题用到的是食物链的思想，以（x+n）代表敌对关系 #include <cstdio> #include <iostream> #include <cmath> #include <string> #include <cstring> #include <algorithm> #include <queue> using namespace std; typedef long long ll; const int inf = 0x3f3f3f3f; const int maxn = 1e6; int sign[maxn]; int n, m, t; void init() { for(int i = 0; i<maxn; i++) sign[i] = i; } int get(int x) { if(x == sign[x]) return x; return sign[x] = get(sign[x]); } void unio(int x, int y)//合并敌对关系 { int a = get(x), b = get(y); sign[a] = b; } int cmp1(int x, int y)//判断两个人是不在同一个帮派 { if(get(x) == get(y+n) || get(y) == get(x+n))//如果两者为敌对关系，则返回1 return 1; return 0; } int cmp2(int x, int y)//判断两人在同一个帮派 { if(get(x) == get(y) || get(x+n) == get(y+n))//如果两个人不是敌对关系 return 1; return 0; } int main() { char bang; int a, b; scanf("%d", &t); for(int i = 0; i<t; i++) { init(); scanf("%d%d", &n, &m); for(int j = 0; j<m; j++) { scanf("%s%d%d",&bang,&a,&b); if(bang == 'D') { unio(a, b + n); unio(a + n,b); } else if(bang == 'A') { if(cmp1(a, b))//如果不在同一个帮派 printf("In different gangs.\n"); else if(cmp2(a,b))//如果在同一个帮派 printf("In the same gang.\n"); else//搜索不到 printf("Not sure yet.\n"); } } } return 0; }

 

转载于:https://www.cnblogs.com/RootVount/p/10539618.html