Dragon Balls

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2273 Accepted Submission(s): 888

Problem Description

Five hundred years later, the number of dragon balls will increase unexpectedly, so it's too difficult for Monkey King(WuKong) to gather all of the dragon balls together.

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.

Input

The 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)

Output

For 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

Case 1: 2 3 0 Case 2: 2 2 1 3 3 2

分析：

本题主要是用到并查集。难点为，怎么保存转移次数。考虑并查集的路径压缩，就是通过路径压缩来更新转移的次数，比如每次移动时，我只需要把这个城市的根结点的转移次数+1，等到以后路径压缩时，子结点自己移动的次数加上根结点移动的次数，就是这个结点总共的移动次数。

代码如下：

1 #include
      
        2 #include
       
         3 #include
        
          4 #include
         
           5 #include
           6 #include
           
             7 #define N 10005 8 using namespace std; 9 struct node10 {11 int parent; // 根节点12 int son; // 子节点13 int rank; // 转移次数14 }p[N];15 int n;16 void init()17 {18 for(int i=1;i<=n;i++)19 {20 p[i].parent = i;21 p[i].rank =0;22 p[i].son = 1;23 }24 }25 int Find(int x)26 {27 if(x== p[x].parent) return x;28 int temp_parent=p[x].parent;29 p[x].parent= Find(p[x].parent);30 p[x].rank+=p[temp_parent].rank; //因为，根节点连到另一个新的根节点后，根节点有变化，故需要更新x到新根节点的转移次数31 return p[x].parent;32 }33 void Union(int x, int y)34 {35 int xf=Find(x);36 int yf=Find(y);37 p[xf].parent=yf;38 p[xf].rank++; 39 p[yf].son+=p[xf].son;40 p[xf].son=0;41 }42 int main()43 {44 int t,count=1;45 scanf("%d",&t);46 while(t--)47 {48 int k,x,y;49 char c;50 printf("Case %d:\n",count++);51 scanf("%d%d",&n,&k);52 init();53 for(int i=0;i

转载于:https://www.cnblogs.com/zn505119020/p/3580055.html

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