Cube Stacking(并差集深度+结点个数)

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 21567		Accepted: 7554
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
题意:有n个从1到n编号的箱子,将每个箱子当做一个栈,对这些箱子进行p次操作,每次操作分别为以下两种之一:
           1>输入 M x y:表示将编号为x的箱子所在的栈放在编号为y的箱子所在栈的栈顶.
           2>输入 C x:计算编号为x的所表示的栈中在x号箱子下面的箱子数目.
题解：和龙珠那题一样，坑了半天，写的都醉了。。。
ac代码贴上：

 #include<stdio.h>
 #include<string.h>
 const int MAXN=;
 int pre[MAXN],s[MAXN],dep[MAXN];//s数组存当前树的节点数；
 int find(int x){
     if(x!=pre[x]){
         int t=pre[x];
         pre[x]=find(pre[x]);
         dep[x]+=dep[t];
     }
 /*    int i=x,j;
     while(i!=r){
         j=pre[i];pre[i]=r;i=j;
     }*/
     return pre[x];
 }
 void merge(int x,int y){
     int f1,f2;
     f1=find(x);f2=find(y);
 //    printf("%d %d\n",f1,f2);
     if(f1!=f2){
         pre[f2]=f1;
         dep[f2]=s[f1];
         s[f1]+=s[f2];
     }
 }
 int main(){
     int N,x,y;
     char c[];//定义成s了，错的啊。。。。
     while(~scanf("%d",&N)){
         for(int i=;i<=N;i++)pre[i]=i,dep[i]=,s[i]=;
         while(N--){
         scanf("%s",c);
         if(c[]=='M'){
             scanf("%d%d",&x,&y);
             merge(x,y);
         }
         else {
             scanf("%d",&x);
             int px=find(x);
         //    printf("s[%d]=%d %d\n",px,s[px],dep[x]);
             printf("%d\n",s[px]-dep[x]-);
             }
         }
     }
     return ;
 }

另一种解法超时：

 #include<stdio.h>
 #include<string.h>
 const int MAXN=;
 int pre[MAXN],s[MAXN];//s数组存当前树的节点数；
 int find(int x){
     while(x!=pre[x])
     x=pre[x];
     return x;
 }
 void merge(int x,int y){
     int f1,f2;
     f1=find(x);f2=find(y);
     if(f1!=f2){
         pre[f2]=f1;
         s[f1]+=s[f2];
     }
 }
 int getdep(int x){
     int s=;
     while(x!=pre[x]){
         x=pre[x];
         s++;
     }
     return s;
 }
 int main(){
     int N,x,y;
     char c[];//定义成s了，错的啊。。。。
     while(~scanf("%d",&N)){
         for(int i=;i<=N;i++)pre[i]=i,s[i]=;
         while(N--){
         scanf("%s",c);
         if(c[]=='M'){
             scanf("%d%d",&x,&y);
             merge(x,y);
         }
         else {
             scanf("%d",&x);
             int px=find(x);
             int d=getdep(x);
             printf("%d\n",s[px]-d-);
             }
         }
     }
     return ;
 }