AOJ 2170 - Marked Ancestor

http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2170

Problem F: Marked Ancestor

You are given a tree T that consists of N nodes. Each node is numbered from 1 to N, and node 1 is always the root node of T. Consider the following two operations on T:

• M v: (Mark) Mark node v.
• Q v: (Query) Print the index of the nearest marked ancestor of node v which is nearest to it. Initially, only the root node is marked.

Your job is to write a program that performs a sequence of these operations on a given tree and calculates the value that each Q operation will print. To avoid too large output file, your program is requested to print the sum of the outputs of all query operations. Note that the judges confirmed that it is possible to calculate every output of query operations in a given sequence.

Input

The first line of the input contains two integers N and Q, which denotes the number of nodes in the tree T and the number of operations, respectively. These numbers meet the following conditions: 1 ≤ N ≤ 100000 and 1 ≤Q ≤ 100000.

The following N - 1 lines describe the configuration of the tree T. Each line contains a single integer pi (i = 2, ... , N), which represents the index of the parent of i-th node.

The next Q lines contain operations in order. Each operation is formatted as "M v" or "Q v", where v is the index of a node.

The last dataset is followed by a line containing two zeros. This line is not a part of any dataset and should not be processed.

Sample Input

6 3
1
1
2
3
3
Q 5
M 3
Q 5
0 0

Output for the Sample Input

4

http://www.hankcs.com/program/cpp/2170-marked-ancestor.html
染色祖先：距离某节点最近的染色节点称为祖先，求染色过程中各节点祖先。
2.4 加工并储存数据的数据结构

并查集
稍有些复杂，需要bfs和两棵树。一棵树用来维护并查集，一棵树用来保存原来的树。bfs用来求解每个节点的祖先。也许看到这里你很好奇：既然节点的祖先是用bfs求出来的，那并查集是干什么吃的？这题的复杂之处在于，它是多用例，如果每染色一个节点就跑一遍bfs，那么会TLE。并查集用来省时间的。

思路是先染色，同时用个栈保留这些操作。接着跑一遍bfs，得到祖先的最终状态，并且将祖先相同的节点放入同一个集合中。然后从栈里面弹出操作将节点染回来，“反染色”的过程中，更新并查集和祖先。

注意两点：
染色节点的祖先是自己
可能存在重复染色，忽略它们

// 并查集相关数据与算法

#define MAX_N 100000 + 16

int parent[MAX_N];

int height[MAX_N];

void init(const int& n)

{

    for (int i = 0; i < n; ++i)

    {

        parent[i] = i;

        height[i] = 0;

    }

}

int find(const int& x)

{

    if (parent[x] == x)

    {

        return x;

    }

    else

    {

        return parent[x] = find(parent[x]);

    }

}

void unite(int x, int y)

{

    x = find(x);

    y = find(y);

    if (x == y)

    {

        return;

    }

    if (height[x] < height[y])

    {

        parent[x] = y;

    }

    else

    {

        parent[y] = x;

        if (height[x] == height[y])

        {

            ++height[x];

        }

    }

}

bool same(const int& x, const int& y)

{

    return find(x) == find(y);

}

// End Of 并查集

// 原始的树用它来描述

vector<int> children[MAX_N];

int parent_tree[MAX_N];

bool marked[MAX_N];

int ancestor[MAX_N];

// 每条指令被拆分为 操作 + 目标

stack<char> operation;

stack<int>  target;

void bfs(int index, int the_ancestor)

{

    queue<int> q_index;

    queue<int> q_ancestor;

    q_index.push(index);

    q_ancestor.push(the_ancestor);

    while (!q_index.empty())

    {

        the_ancestor = q_ancestor.front(); q_ancestor.pop();

        index = q_index.front(); q_index.pop();

        if (marked[index] == true)

        {

            the_ancestor = index;

        }

        ancestor[index] = the_ancestor;

        for (vector<int>::iterator it = children[index].begin(); it != children[index].end(); ++it)

        {

            q_index.push(*it);

            q_ancestor.push(the_ancestor);

        }

    }

}

///////////////////////////SubMain//////////////////////////////////

int main(int argc, char *argv[])

{

    int N, Q;

    while (cin >> N, cin >> Q, N)

    {

        for (int i = 0; i < N; ++i)

        {

            children[i].clear();

            marked[i] = false;

        }

        marked[0] = true;

        int p;

        for (int i = 1; i < N; ++i)

        {

            cin >> p; --p;

            parent_tree[i] = p;

            children[p].push_back(i);

        }

        for (int i = 0; i < Q; ++i)

        {

            char o;

            int t;

            cin >> o >> t; --t;

            if (o == 'M')

            {

                if (marked[t])

                {

                    continue;

                }

                else

                {

                    marked[t] = true;

                }

            }

            operation.push(o);

            target.push(t);

        }

        bfs(0, 0);

        init(N);

        for (int i = 0; i < N; ++i)

        {

            unite(i, ancestor[i]);

        }

        unsigned long long result = 0;

        while (!operation.empty())

        {

            char o = operation.top(); operation.pop();

            int t = target.top(); target.pop();

            if (o == 'Q')

            {

                result += ancestor[find(t)] + 1;    // 题目index从1开始

            }

            else

            {

                // 执行“反染色”操作，之后这个节点的祖先变为其父节点的祖先

                int p = ancestor[find(parent_tree[t])];

                unite(t, parent_tree[t]);

                ancestor[find(t)] = p;

            }

        }

        cout << result << endl;

    }

     

    return 0;

}

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