HOJ 2385 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
题解：
并查集，并记录i元素上面和下面的元素个数
#include <cstdio>
#include <cstdlib>
#include <cstring>
const int N = 30005;
int set[N];
int high[N]; //i元素上面（加上i）的元素个数
int low[N]; //i元素下面的元素个数
int findset(int x){
if(x == set[x])
return set[x];
int tm = set[x];
set[x] = findset(set[x]);
low[x] += low[tm];
return set[x];
}
void unionset(int x,int y){
int fx = findset(x);
int fy = findset(y);
if( fx == fy)
return ;
set[fx] = fy;
low[fx] += high[fy];
high[fy] += high[fx];
}
int main(){
int n;
while(scanf("%d",&n) != EOF){
for(int i = 1; i <= N; i++){
set[i] = i;
high[i] = 1;
}
memset(low,0,sizeof(low));
char op;
int x,y;
getchar();
for(int i =1 ; i <= n; i++){
scanf("%c",&op);
if(op == 'C'){
scanf("%d",&x);
findset(x);
printf("%d\n",low[x]);
}
else{
scanf("%d%d",&x,&y);
unionset(x,y);
}
getchar();
}
}
return 0;
}
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