HDU 1988 cube stacking (Data Structure-query set)

Cube stacking

Time limit:2000 ms		Memory limit:30000 K
Total submissions:18900		Accepted:6568
Case time limit:1000 ms

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

* 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

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

Sample output

102

Source

Usaco 2004 u s open

Question:

There are n cubes and N grids, 1 ~ N number. At first, cube I was on the grid I, and each grid had exactly one cube. Now M group operations, m a B indicates that all the cubes in the grid where cube A is located are placed on all the cubes in the grid where cube B is located. C X indicates the number of cubes under cube X.

Solution:

Based on the query, you only need to know the distance from X to the father and the distance from the father to the end to know the distance from X to the end. The distance between the sum [I] record and the root is maintained continuously.

Reference code:

#include <iostream>#include <vector>#include <map>using namespace std;const int MAXN = 30010;int p;int father[MAXN], sum[MAXN], num[MAXN];void init() {    for (int i = 1; i <= 30000; i++) {        father[i] = i;        sum[i] = 0;        num[i] = 1;    }}int find_set(int x) {    int tmp = father[x];    if (father[x] != x) {        father[x] = find_set(father[x]);        sum[x] += sum[tmp];    }    return father[x];}void union_set(int x, int y) {    x = find_set(x);    y = find_set(y);    if (x == y) return;    father[x] = y;    sum[x] += num[y];    num[y] += num[x];}void solve() {    for (int k = 0; k < p; k++) {        char op;        cin >> op;        if (op == 'M') {            int x, y, posx;            cin >> x >> y;            union_set(x, y);        } else if (op == 'C') {            int x, cnt = 0;            cin >> x;            find_set(x);            cout << sum[x] << endl;        }    }}int main() {    ios::sync_with_stdio(false);    while (cin >> p) {        init();        solve();    }    return 0;}