poj1988_并查集

Cube Stacking

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
记录容器中多少个元素,容器头部多少个元素,减一下;

#include<cstdio>const int N=30030;int p[N],cnt[N],top[N];int Find(int x){    int t;    if(p[x]!=x)    {         t=p[x];        p[x]=Find(p[x]);        top[x]+=top[t];    }    return p[x];}void init(int x,int y){    int tx=Find(x);    int ty=Find(y);    if(tx!=ty)    {        p[ty]=tx;        top[ty]=cnt[tx];        cnt[tx]+=cnt[ty];    }}int main(){    int t,i,j;    for(i=0;i<N;i++)    {        p[i]=i;        cnt[i]=1;        top[i]=0;    }    scanf("%d",&t);    while(t--)    {        char str[10];        int a,b;        scanf("%s",str);        if(str[0]=='M')        {            scanf("%d%d",&a,&b);            init(a,b);        }        else        {            scanf("%d",&a);            int tt=Find(a);            printf("%d\n",cnt[tt]-top[a]-1);        }    }    return 0;}