His country has N cities and there are exactly N dragon balls in the world. At first, for the ith dragon ball, the sacred dragon will puts it in the ith city. Through long years, some cities' dragon ball(s) would be transported to other cities. To save physical strength WuKong plans to take Flying Nimbus Cloud, a magical flying cloud to gather dragon balls.
Every time WuKong will collect the information of one dragon ball, he will ask you the information of that ball. You must tell him which city the ball is located and how many dragon balls are there in that city, you also need to tell him how many times the ball has been transported so far.
InputThe first line of the input is a single positive integer T(0 < T <= 100).
For each case, the first line contains two integers: N and Q (2 < N <= 10000 , 2 < Q <= 10000).
Each of the following Q lines contains either a fact or a question as the follow format:
T A B : All the dragon balls which are in the same city with A have been transported to the city the Bth ball in. You can assume that the two cities are different.
Q A : WuKong want to know X (the id of the city Ath ball is in), Y (the count of balls in Xth city) and Z (the tranporting times of the Ath ball). (1 <= A, B <= N)OutputFor each test case, output the test case number formated as sample output. Then for each query, output a line with three integers X Y Z saparated by a blank space.Sample Input
2 3 3 T 1 2 T 3 2 Q 2 3 4 T 1 2 Q 1 T 1 3 Q 1
Sample Output
. 1 Case: 2. 3 0 Case 2: 2 2. 1 . 3. 3 2
disjoint-set with the right
// This question is feeling a pit ,,, is when a city of Pearl is moved away, then this city is useless, there will be no movement over the Pearl // because the TAB (A and B are Dragon Ball) , so the root node on the Pearl index corresponds to his city, // so finding one city where we just Dragonball requires that its root just fine. Seeking a number of cities in the Pearl, // nothing more than to find the root of this tree the number of child nodes (and he is considered). The main demand is the number of Dragon Ball movement. // we opened a record array, initialized to 0 when connecting a two Pearl, we connect these two Pearl root node // and then we let the root move, and then let the root node on a mobile node so ,,, ,,. #include <cstdio> #include <CString> the using namespace STD; const int N = 1E5 + . 7 ; int FA [N]; // record the parent node int Son [N]; // record the size tree int RAN [N]; // record number of moves int Find ( int X) { IF (X == FA [X]) return FA [X]; the else { int K = FA [X]; FA [X] = Find (FA [X]); RAN [X] RAN = + [K]; // this is mainly used to transmit movement of the root node. return FA [X]; } } void the Join ( int X, int Y) { int FX = Find (X), FY = Find (Y); IF (! FX = FY) { FA [FX] = FY; Son [ FY] + = Son [FX]; RAN [FX] ++; // let the mobile root } } // Initialization void inint ( int X) { Memset (RAN, 0 , the sizeof (RAN)); for ( int I = . 1 ; I <= X; ++ I ) { FA [I] = I; Son [I] = . 1 ; } } int main () { int T, KK = 0 ; Scanf ( " % D " , & T); the while (T-- ) { KK ++ ; printf("Case %d:\n",kk); int n,m; scanf("%d%d",&n,&m); inint(n); for(int i=1;i<=m;i++){ char a[10]; scanf("%s",a); if(a[0]=='T'){ int x,y; scanf("%d%d",&x,&Y); join(x,y); } else if(a[0]=='Q'){ int z; scanf("%d",&z); int m=find(z); printf("%d %d %d\n",m,son[m],ran[z]); } } } return 0; }