Cube StackingDescription
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
6M 1 6C 1M 2 4M 2 6C 3C 4Sample Output
102Source
【题意】
约翰和贝西在玩叠积木，每块积木是大小一样的立方体，将几块积木叠在一起可以得到更高的长 方体。积木是有磁性的，叠起来的积木分不开。 约翰负责把积木叠高，而贝西负责清点积木的高度。一开始共有 N 块积木，编号为 1 到 N，所 有积木都是分开的，没有积木叠在一起。 每次移动积木的时候，约翰会先选择一块积木 X，再选择另一块积木 Y，把 X 搬到 Y 的上方。 如果 X 已经和其它积木叠在一起了，那么应将这叠积木整体移动到 Y 的上方；如果 Y 已经和其它积 木叠在一起了的，假设在 Y 上方最高处的积木为 Z，那么应将 X 所在的那叠积木移动到 Z 的上方。 在约翰辛苦劳动的同时，贝西在一边计算积木已经叠得多高了。约翰和贝西是交替工作的，可惜 贝西不太擅长于数数，请你帮她计算一下，在一些时刻，指定的积木的下方还有多少其他积木。
【分析】
我这个并查集的权值是他跟父亲的距离，最后做一遍find就知道他跟最低点的距离（即下面有多少个东西），然后我还开多了一个并查集，表示这个东西最上面是什么东西。
差不多这样把。
代码如下：
1 #include2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 using namespace std; 9 #define Maxn 30001010 11 int fa[Maxn],d[Maxn],g[Maxn];12 char s[10];13 14 int ffind(int x)15 {16 int y=fa[x];17 if(fa[x]!=x) fa[x]=ffind(fa[x]);18 d[x]+=d[y];19 return fa[x];20 }21 22 int find_g(int x)23 {24 if(g[x]!=x) g[x]=find_g(g[x]);25 return g[x];26 }27 28 int main()29 {30 int q;31 scanf("%d",&q);32 for(int i=1;i<=300010;i++) fa[i]=i,g[i]=i,d[i]=0;33 for(int i=1;i<=q;i++)34 {35 scanf("%s",s);36 if(s[0]=='M')37 {38 int x,y;39 scanf("%d%d",&x,&y);40 int z=ffind(x);y=find_g(y);41 fa[z]=y;d[z]=1;42 g[find_g(y)]=find_g(x);43 }44 else45 {46 int x;47 scanf("%d",&x);48 ffind(x);49 printf("%d\n",d[x]);50 }51 }52 return 0;53 }
2016-10-27 20:25:21