Cube Stacking

Cube Stacking
Time Limit: 2000MS Memory Limit: 30000K
Total Submissions: 21350 Accepted: 7470
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
并查集的合并,每个点记录到栈底的距离,栈底的元素记录栈的大小方便合并

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define LL long longusing namespace std;const int MAX =  31000;const int INF = 0x3f3f3f3f;struct node{    int pre;    int dis;    int Size;}a[MAX];int m;int Find(int x){    if(a[x].pre==x)    {        return x;    }    int t=a[x].pre;    a[x].pre=Find(t);    a[x].dis+=a[t].dis;//将距离更新    return a[x].pre;}void Join(int x,int y){    int b=Find(x);    int c=Find(y);    a[b].pre=c;//将两个栈的栈底相连    a[b].dis+=a[c].Size;//上面的栈底更新距离    a[c].Size+=a[b].Size;//下面的栈底更新大小    a[b].Size=0;//不在为栈底,大小为零}int main(){    scanf("%d",&m);    char s[5];    int u,v;    for(int i=0;i<MAX;i++)    {        a[i].dis=0;        a[i].Size=1;        a[i].pre=i;    }    for(int i=0;i<m;i++)    {        scanf("%s",s);        if(s[0]=='M')        {            scanf("%d %d",&u,&v);            Join(u,v);        }        else        {            scanf("%d",&u);            Find(u);//要先更新一遍            printf("%d\n",a[u].dis);        }    }    return 0;}