• ACM学习历程——POJ3321 Apple Tree(搜索,线段树)


         

    Description

    There is an apple tree outside of kaka's house. Every autumn, a lot of apples will grow in the tree. Kaka likes apple very much, so he has been carefully nurturing the big apple tree.

    The tree has N forks which are connected by branches. Kaka numbers the forks by 1 to N and the root is always numbered by 1. Apples will grow on the forks and two apple won't grow on the same fork. kaka wants to know how many apples are there in a sub-tree, for his study of the produce ability of the apple tree.

    The trouble is that a new apple may grow on an empty fork some time and kaka may pick an apple from the tree for his dessert. Can you help kaka?

    Input

    The first line contains an integer N (N ≤ 100,000) , which is the number of the forks in the tree. The following N - 1 lines each contain two integers u and v, which means fork u and fork v are connected by a branch. The next line contains an integer M (M ≤ 100,000). The following M lines each contain a message which is either "C x" which means the existence of the apple on fork x has been changed. i.e. if there is an apple on the fork, then Kaka pick it; otherwise a new apple has grown on the empty fork. or "Q x" which means an inquiry for the number of apples in the sub-tree above the fork x, including the apple (if exists) on the fork x Note the tree is full of apples at the beginning

    Output

    For every inquiry, output the correspond answer per line.

    Sample Input

    3
    1 2
    1 3
    3
    Q 1
    C 2
    Q 1
    

    Sample Output

    3
    2
    

    这道题的关键是如何把这题转换成一个区间操作的问题。

    由于每次查询的是子树中苹果的个数,说明查询的区间是子树的中节点的所在区间,即子树中所有节点应在查区间内。即要创建一种映射,使得,子树中节点的标号在查询区间内。

    于是一种映射便是,对每个结点赋予两个值lt和rt,lt表示以此点为根节点的子树中序号最小的,rt表示当前节点的序号。

    这种映射可以通过Dfs生成。然后题目就转换成了区间操作的问题。

    不过对于存初始图的问题,我一开始使用了STL里的vector和list都超时了。最后用了链式前向星。

    代码:

    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    #include <set>
    #include <map>
    #include <vector>
    #include <list>
    #include <queue>
    #include <string>
    #define inf 0xfffffff
    #define eps 1e-10
    #define N 1000000007
    
    using namespace std;
    
    //线段树
    //区间每点增值,求区间和
    const int maxn = 100005;
    struct node
    {
        int lt, rt;
        int val;
    }tree[4*maxn];
    
    //向上更新
    void PushUp(int id)
    {
        tree[id].val = tree[id<<1].val + tree[id<<1|1].val;
    }
    
    //建立线段树
    void Build(int lt, int rt, int id)
    {
        tree[id].lt = lt;
        tree[id].rt = rt;
        tree[id].val = 0;//每段的初值,根据题目要求
        if (lt == rt)
        {
            tree[id].val = 1;
            return;
        }
        int mid = (lt + rt) >> 1;
        Build(lt, mid, id<<1);
        Build(mid + 1, rt, id<<1|1);
        PushUp(id);
    }
    
    //增加区间内每个点固定的值
    void Add(int lt, int rt, int id, int pls)
    {
        if (lt <= tree[id].lt && rt >= tree[id].rt)
        {
            if (tree[id].val)
                pls = -pls;
            tree[id].val += pls * (tree[id].rt-tree[id].lt+1);
            return;
        }
        int mid = (tree[id].lt + tree[id].rt) >> 1;
        if (lt <= mid)
            Add(lt, rt, id<<1, pls);
        if (rt > mid)
            Add(lt, rt, id<<1|1, pls);
        PushUp(id);
    }
    
    //查询某段区间内的和
    int Query(int lt, int rt, int id)
    {
        if (lt <= tree[id].lt && rt >= tree[id].rt)
            return tree[id].val;
        int mid = (tree[id].lt + tree[id].rt) >> 1;
        int ans = 0;
        if (lt <= mid)
            ans += Query(lt, rt, id<<1);
        if (rt > mid)
            ans += Query(lt, rt, id<<1|1);
        return ans;
    }
    
    //链式前向星
    struct Edge
    {
        int to, next;
    }edge[200005];
    
    int head[100005], cnt;
    
    void AddEdge(int u, int v)
    {
        edge[cnt].to = v;
        edge[cnt].next = head[u];
        head[u] = cnt;
        cnt++;
    }
    
    void InitEdge()
    {
        memset(head, -1, sizeof(head));
        cnt = 0;
    }
    
    struct
    {
        int lt, rt;
    }id[100005];
    
    int n, m, now;
    
    
    void Dfs(int k)
    {
        if (id[k].lt != 0)
            return;
        id[k].lt = now;
        for (int i = head[k]; i != -1; i = edge[i].next)
        {
            Dfs(edge[i].to);
        }
        id[k].rt = now;
        now++;
    }
    
    void Init()
    {
        memset(id, 0, sizeof(id));
        int u, v;
        InitEdge();
        for (int i = 1; i < n; ++i)
        {
            scanf("%d%d", &u, &v);
            AddEdge(u, v);
            AddEdge(v, u);
        }
        now = 1;
        Dfs(1);
        Build(1, n, 1);
    }
    
    void Work()
    {
        char op[3];
        int v;
        for (int i = 0; i < m; ++i)
        {
            scanf("%s%d", op, &v);
            if (op[0] == 'C')
            {
                Add(id[v].rt, id[v].rt, 1, 1);
            }
            else
            {
                printf("%d
    ", Query(id[v].lt, id[v].rt, 1));
            }
        }
    }
    
    int main()
    {
        //freopen("test.in", "r", stdin);
        while (scanf("%d", &n) != EOF && n)
        {
            Init();
            scanf("%d", &m);
            Work();
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/andyqsmart/p/4469696.html
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