HDOJ 2818 Building Block

Building Block

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4222 Accepted Submission(s): 1316



Problem Description

John are playing with blocks. There are N blocks (1 <= N <= 30000) numbered 1...N。Initially, there are N piles, and each pile contains one block. Then John do some operations P times (1 <= P <= 1000000). There are two kinds of operation:

M X Y : Put the whole pile containing block X up to the pile containing Y. If X and Y are in the same pile, just ignore this command.

C X : Count the number of blocks under block X

You are request to find out the output for each C operation.

Input

The first line contains integer P. Then P lines follow, each of which contain an operation describe above.

Output

Output the count for each C operations in one line.

Sample Input

6
M 1 6
C 1
M 2 4
M 2 6
C 3
C 4

Sample Output

1
0
2

#include <cstdio>
#include <cstring>
using namespace std;

int father[30005];
int rank[30005];
int maxNum[30005];

int find(int x){
if (father[x] != x){
int fa = father[x];
father[x] = find(fa);
rank[x] += rank[fa];
}
return father[x];
}

void Union(int x, int y){
int fx = find(x);
int fy = find(y);
if (fx != fy){
father[fx] = fy;
rank[fx] += maxNum[fy] + 1;
maxNum[fy] += maxNum[fx] + 1;
}
}

int main(){
int t;
int x, y, i;
char c;
while (scanf("%d%*c", &t) != EOF){
memset(rank, 0, sizeof(rank));
memset(maxNum, 0, sizeof(maxNum));

for (i = 0; i <= 30005; i++)
father[i] = i;
while (t--){
c = getchar();
if (c == 'M'){
scanf("%d%d%*c", &x, &y);
Union(x, y);
}
else{
scanf("%d%*c", &x);
find(x);
printf("%d\n", rank[x]);
}
}
}
return 0;
}