poj 1988 Cube Stacking

Cube Stacking

Time Limit: 2000MS Memory Limit: 30000KTotal Submissions: 16031 Accepted: 5477Case 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

注意合并时对根节点的处理。

AC代码：

#include <cstdio>#include <cstring>#define N 30005using namespace std;int father[N],up[N],num[N];   //up[i]为i上位的方块，num[i]为i所在堆的总方块数int Find_set(int n){    int temp;    if(n==father[n]) return n;    temp=father[n];    father[n]=Find_set(father[n]);    up[n]=up[n]+up[temp];            //更新n的偏移量，n的新偏移量等于n相对旧的根结点的偏移量+旧的根节点相对新的根节点的偏移量    return father[n];}void Union(int a,int b){    int ra=Find_set(a);    int rb=Find_set(b);    if(ra==rb) return;    father[rb]=ra;                  //将ra所在堆放在rb所在堆上面    up[rb]+=num[ra];                //rb上面的数目要加上ra所在堆的数目    num[ra]+=num[rb];               //a所在堆的数目要加上rb所在堆的数目}int main(){    int p,a,b;    char s[3];    while(scanf("%d",&p)!=EOF)    {        for(int i=0; i<N; i++)        {            father[i]=i;            num[i]=1;            up[i]=0;        }        while(p--)        {            scanf("%s",s);            if(s[0]=='M')            {                scanf("%d%d",&a,&b);                Union(a,b);            }            else            {                scanf("%d",&a);                int root=Find_set(a);                printf("%d\n",num[root]-up[a]-1);            }        }    }    return 0;}