Cube Stacking
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 24665 | Accepted: 8653 | |
| 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
6M 1 6C 1M 2 4M 2 6C 3C 4
Sample Output
102
Source
//同P2342 叠积木#includeusing namespace std;const int N=3e5+5;int n,fa[N],top[N],cnt[N];int find(int x){ if(fa[x]==x) return x; int t=fa[x]; fa[x]=find(fa[x]); fa[x]=fa[t]; top[x]=top[t]; cnt[x]=cnt[t]+cnt[x]; return fa[x];}int main(){ scanf("%d",&n);char s[10]; for(int i=1;i<=30000;i++) fa[i]=top[i]=i; for(int i=1,a,b,x,y;i<=n;i++){ scanf("%s",s); if(s[0]=='M'){ scanf("%d%d",&a,&b); x=find(a);y=find(b); fa[x]=y;find(top[y]); cnt[x]=cnt[top[y]]+1; top[y]=top[x]; } else{ scanf("%d",&x);find(x); printf("%d\n",cnt[x]); } } return 0;}