Poj 1988 Cube Stacking(带权并查集)

Cube Stacking
Time Limit: 2000MS Memory Limit: 30000K
Total Submissions: 24448 Accepted: 8560
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
Source
USACO 2004 U S Open

/*带权并查集.sum[i]表示以i为根的集合的元素个数.cnt[i]表示i元素到祖宗节点的距离.father[i]表示路径压缩后i的祖宗.然后查的时候只需要知道这个块在哪个正方体里和它在这个正方体里的相对位置就可以了. */#include<iostream>#include<cstdio>#define MAXN 30001using namespace std;int n,m,father[MAXN],sum[MAXN],top[MAXN],cnt[MAXN];inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9') x=x*10+ch-48,ch=getchar();    return x*f;}inline int find(int x){    if(x==father[x]) return x;    int fa=father[x];    father[x]=find(father[x]);    cnt[x]+=cnt[fa];    return father[x];}inline int merge(int x,int y){    father[y]=x;    cnt[y]+=sum[x];    sum[x]+=sum[y];    sum[y]=0;}inline int abs(int x){    return x<0?-x:x;}int main(){    int x,y;    n=read();    for(int i=1;i<=MAXN;i++) father[i]=i,sum[i]=1;    char ch[4];    while(n--)    {        cin>>ch+1;        if(ch[1]=='M')         {            x=read(),y=read();            int l1=find(x),l2=find(y);            merge(l1,l2);        }        else {            x=read();int l1=find(x);            printf("%d\n",sum[l1]-cnt[x]-1);        }       }    return 0;}