POJ1988 并查集的使用

Cube Stacking

Time Limit: 2000MS   Memory Limit: 30000K
Total Submissions: 21157   Accepted: 7395
Case Time Limit: 1000MS

Description

Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations:
moves and counts.
* In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y.
* In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value.

Write a program that can verify the results of the game.

Input

* Line 1: A single integer, P

* Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a 'M' for a move operation or a 'C' for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X.

Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself.

Output

Print the output from each of the count operations in the same order as the input file.

Sample Input

6
M 1 6
C 1
M 2 4
M 2 6
C 3
C 4

Sample Output

1
0
2

 1 #include<stdio.h>
 2 #define N 30001
 3 
 4 int count[N], num[N], pre[N];
 5 
 6 void inite()
 7 {
 8     for(int i = 0; i < N; i++)
 9     {
10         count[i] = 0;
11         num[i] = 1;
12         pre[i] = i;
13     }
14 }
15 
16 int find(int x)
17 {
18     if(pre[x] == x)
19         return x;
20 
21     int t = find(pre[x]);
22     count[x] += count[pre[x]];
23     pre[x] = t;
24     return t;
25 
26 }
27 void Union(int x, int y)
28 {
29     int i = find(x);
30     int j = find(y);
31     if(i == j)
32     {
33         return;
34     }
35     count[i] = num[j];
36     num[j] += num[i];
37     pre[i] = j;
38 }
39 
40 
41 
42 int main()
43 {
44     int i, x, y, n;
45     char s[2];
46     scanf("%d",&n);
47     inite();
48     for(i = 0; i < n; i++)
49     {
50         scanf("%s",s);
51         if(s[0] == 'M')
52         {
53             scanf("%d%d",&x,&y);
54             Union(x,y);
55         }
56         else if(s[0] == 'C')
57         {
58             scanf("%d",&x);
59             int c = find(x);
60             printf("%d
",count[x]);
61         }
62     }
63     return 0;
64 }
View Code
 
原文地址:https://www.cnblogs.com/cjshuang/p/4678789.html