Cube Stacking
Time Limit: 2000MS Memory Limit: 30000K
Total Submissions: 22236 Accepted: 7803
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
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
题意:初始时有n个堆栈,每个堆栈上面1个cube。(编号都为1 ~ n),现在定义下面两个操作:
操作’M’ X Y: 按数据结构的栈来操作,将含有X元素的栈的元素,放到元素Y的上面
操作’C’ X:统计第X个cube下面含有多少个cube
思路:分别用under[i]保存第i个cube下面的cube数目;num[x]表示根为x的集合元素数;fat[x]表示元素x所在集合的根。
例如:M X Y的操作变化如下
#include
#include
#include
#include
#include
using namespace std;
const int maxn = 30005;
int fat[maxn],P,x,y,num[maxn],under[maxn];
char m[10];
int find(int x){
int tmp=fat[x];
if(x!=fat[x])
fat[x]=find(fat[x]),under[x]+=under[tmp];//为什么这里要加上under[tmp],见图一
return fat[x];
}
void Union(int x,int y){
int fx=find(x),fy=find(y);
if(fx==fy) return;
fat[fx]=fy;
under[fx]+=num[fy];
num[fy]+=num[fx];
}
int main(){
freopen("i.txt","r",stdin);
cin>>P;
for(int i=0;i0,num[i]=1;
while(P--){
scanf("%s",m);
if(m[0]=='M'){
scanf("%d%d",&x,&y);
Union(x,y);
}
else{
scanf("%d",&x);
find(x);
cout<return 0;
}