ZOJ3696-A Simple Tree Problem 线段树

最新推荐文章于 2018-04-07 10:20:45 发布

原创最新推荐文章于 2018-04-07 10:20:45 发布 · 711 阅读

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本文介绍了一种解决二叉树操作和查询问题的高效算法，包括节点的布尔标签修改和子树中1的数量统计。通过深度优先搜索预处理构建线段树，实现了快速的操作和查询。实例演示了水题的解决方法，适用于初学者理解二叉树的高级应用。

A Simple Tree Problem

Time Limit: 3 Seconds Memory Limit: 65536 KB

Given a rooted tree, each node has a boolean (0 or 1) labeled on it. Initially, all the labels are 0.

We define this kind of operation: given a subtree, negate all its labels.

And we want to query the numbers of 1's of a subtree.

Input

Multiple test cases.

First line, two integer N and M, denoting the numbers of nodes and numbers of operations and queries.(1<=N<=100000, 1<=M<=10000)

Then a line with N-1 integers, denoting the parent of node 2..N. Root is node 1.

Then M lines, each line are in the format "o node" or "q node", denoting we want to operate or query on the subtree with root of a certain node.

Output

For each query, output an integer in a line.

Output a blank line after each test case.

Sample Input

Sample Output

水题，先dfs，再线段树

/* 
 * File:   main.cpp
 * Author: Kevin
 *
 * Created on 2013年10月28日, 下午3:56
 */

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<utility>
using namespace std;
#define MAXN 100005
#define lson(x) (x<<1)
#define rson(x) (x<<1|1)
#define pret(x) (x>>1)
#define lens(x) (cord[x].second-cord[x].first+1)
typedef pair<int, int> pr;
vector<int>vt[MAXN];
pr mk[MAXN];
/**************线段树相关数组***************/
pr cord[MAXN<<2]; //区间端点
int sum[MAXN<<2]; //所求答案
int flag[MAXN<<2]; //延迟标记
/********线段树的最大值应该是点数4倍**********/
int id;
void dfs(int rt) {
    mk[rt].first = ++id;
    for (int i = 0; i < vt[rt].size(); i++) {
        int v = vt[rt][i];
        if (mk[v].first == 0)dfs(v);
    }
    mk[rt].second = id;
}

void pushup(int rt) {
    sum[rt] = sum[lson(rt)] + sum[rson(rt)];
}

void pushdown(int rt) {
    if (flag[rt]) {
        flag[rt] = 0;
        flag[lson(rt)] ^= 1;
        flag[rson(rt)] ^= 1;
        sum[lson(rt)] = lens(lson(rt)) - sum[lson(rt)];
        sum[rson(rt)] = lens(rson(rt)) - sum[rson(rt)];
    }
}

void build(int l, int r, int rt) {
    cord[rt] = make_pair(l, r);
    flag[rt] = sum[rt] = 0;
    if (l == r)return;
    int m = pret(l + r);
    build(l, m, lson(rt));
    build(m + 1, r, rson(rt));
}

void update(int l, int r, int rt) {
    if (cord[rt].first >= l && cord[rt].second <= r) {
        flag[rt] ^= 1;
        sum[rt] = lens(rt) - sum[rt];
        return;
    }
    pushdown(rt);
    int m = pret(cord[rt].first + cord[rt].second);
    if (l <= m)update(l, r, lson(rt));
    if (r >= m + 1)update(l, r, rson(rt));
    pushup(rt);
}

int query(int l, int r, int rt) {
    if (cord[rt].first >= l && cord[rt].second <= r) {
        return sum[rt];
    }
    int ret = 0;
    int m = pret(cord[rt].first + cord[rt].second);
    pushdown(rt);
    if (l <= m)ret += query(l, r, lson(rt));
    if (r >= m + 1)ret += query(l, r, rson(rt));
    return ret;
}

int main(int argc, char** argv) {
    int n, m;
    while (~scanf("%d%d", &n, &m)) {
        for (int i = 1; i <= n; i++)vt[i].clear();
        for (int i = 2; i <= n; i++) {
            int x;
            scanf("%d", &x);
            vt[x].push_back(i);
        }
        for (int i = 1; i <= n; i++)mk[i].first = 0;
        id = 0;
        dfs(1);
        build(1, n, 1);
        while (m--) {
            char c; int x;
            scanf("%*c%c%d", &c, &x);
            if (c == 'o') 
                update(mk[x].first, mk[x].second, 1);
            else 
                printf("%d\n", query(mk[x].first, mk[x].second, 1));
        }
        printf("\n");
    }
    return 0;
}