Building Block[HDU2818]

Building Block

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5426 Accepted Submission(s): 1663

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

路徑壓縮的並查集。對每個點x，rank[x]記錄到父親的距離,size[x]表示以x為根節點的子樹大小。x下邊的積木數=x的最高級父親father的size-Σ(從x到father的路徑上各點的rank)。路徑壓縮這裏寫的有點意識模糊，居然一遍AC，提交完自己都挺奇怪，這就過了啊......

#include <stdio.h>
#include <string.h>
int tmpx;
int size[30005], rank[30005];
class Union_Find_Set {
#define MAX_UNION_FIND_SET_SIZE 30005
public:
    int setSize;
    int father[MAX_UNION_FIND_SET_SIZE];
    Union_Find_Set() {
        setSize = 0;
    }
    Union_Find_Set(int x) {
        setSize = x;
        clear(x);
    }
    void clear(int x) {
        for (int i = 0; i < x; i++) {
            father[i] = i;
        }
    }
    int getFather(int x) {
        tmpx = 0;
        int ret = x, tmp;
        while (ret != father[ret]) {
            tmpx += (rank[ret]);
            ret = father[ret];
        }
        int tmpz = tmpx;
        while (x != father[x]) {
            tmp = father[x];
            father[x] = ret;
            tmpz -= rank[x];
            rank[x] += tmpz;
            x = tmp;
        }
        return ret;
    }
    bool merge(int a, int b) {
        a = getFather(a);
        b = getFather(b);
        if (a != b) {
            father[b] = a;
            rank[b] = size[a] + 1;
            size[a] += (size[b] + 1);
            return true;
        } else {
            return false;
        }
    }
    int countRoot() {
        int ret = 0;
        for (int i = 0; i < setSize; i++) {
            if (father[i] = i) {
                ret++;
            }
        }
        return ret;
    }
};

Union_Find_Set ufs;
int main() {
    int n, a, b;
    char q;
    while (scanf("%d", &n) != EOF) {
        memset(size, 0, sizeof(size));
        memset(rank, 0, sizeof(rank));
        ufs.clear(30001);
        while (n--) {
            getchar();
            scanf("%c", &q);
            if (q == 'M') {
                scanf("%d%d", &a, &b);
                ufs.merge(a, b);
            } else if (q == 'C') {
                scanf("%d", &a);
                printf("%d\n", size[ufs.getFather(a)] - tmpx);
            }
        }
    }
    return 0;
}

文章來源：