Codeforces Round #287 (Div. 2) C. Guess Your Way Out!（线段树思想）

题意:

给定一个深度H≤50个完全二叉树,将叶子节点标号1∼2H
给定移动序列“LRLRLRLR...”,现在要从根到N号叶子
1.L表示去左儿子,R去右儿子
2.如果这个儿子走过了跳过这次指令
3.两个儿子都走过了就回到父亲
4.到叶子但是不是终点也回到父亲
5.到N号叶子游戏结束
问走了多少步

分析:

使用线段树的思想模拟,如果当前移动方向与线段树区间的方向相同那么ans+=1,否则ans+=2∗区间长度

代码:

//
// Created by TaoSama on 2016-02-12
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

typedef long long LL;
LL h, n, ans;

//0->left 1->right
void dfs(LL o, bool dir, LL l, LL r) {
// pr(l);prln(r);
    if(l == r) return;
    LL m = l + r >> 1;
    if(o <= m) {
        if(dir == 0) ++ans;
        else ans += r - m << 1;
        dfs(o, 1, l, m);
    } else {
        if(dir == 0) ans += m + 1 - l << 1;
        else ++ans;
        dfs(o, 0, m + 1, r);
    }
}

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(cin >> h >> n) {
        ans = 0;
        dfs(n, 0, 1, 1LL << h);
        cout << ans << endl;
    }
    return 0;
}