POJ 1988 Cube Stacking

题意：有编号为1~N的N个小木块，有两种操作

1. M x y 将木块x所在的堆放到木块y所在的堆的上面
2. C x 询问木块x下面有多少块木块

代码巧妙就巧妙在GetParent函数中在进行路径压缩的同时，也计算好了该木块对应的under值

这个需要好好体会

 

 1 //#define LOCAL

 2 #include <iostream>

 3 #include <cstdio>

 4 #include <cstring>

 5 using namespace std;

 6 

 7 const int maxn = 30000 + 10;

 8 int parent[maxn], under[maxn], sum[maxn];

 9 

10 int GetParent(int a)

11 {

12     if(parent[a] == a)

13         return a;

14     int t = GetParent(parent[a]);

15     under[a] += under[parent[a]];

16     parent[a] = t;

17     return t;

18 }

19 

20 void Merge(int a, int b)

21 {

22     int pa = GetParent(a);

23     int pb = GetParent(b);

24     if(pa == pb)    return;

25     parent[pb] = pa;

26     under[pb] = sum[pa];

27     sum[pa] += sum[pb];

28 }

29 

30 int main(void)

31 {

32     #ifdef LOCAL

33         freopen("1988in.txt", "r", stdin);

34     #endif

35 

36     for(int i = 1; i < maxn; ++i)

37     {

38         parent[i] = i;

39         under[i] = 0;

40         sum[i] = 1;

41     }

42     int n;

43     scanf("%d", &n);

44     getchar();

45     for(int i = 0; i < n; ++i)

46     {

47         char p;

48         scanf("%c", &p);

49         if(p == 'M')

50         {

51             int a, b;

52             scanf("%d%d", &a, &b);

53             getchar(); 

54             Merge(b, a);

55         }

56         else

57         {

58             int a;

59             scanf("%d", &a);

60             getchar();

61             GetParent(a);

62             printf("%d\n", under[a]);

63         }

64     }

65     return 0;

66 }

代码君