【codeforces 792D】Paths in a Complete Binary Tree

【题目链接】:http://codeforces.com/contest/792/problem/D

【题意】



给你一棵满二叉树;

给你初始节点;

给你若干个往上走,左走,右走操作;

让你输出一系列操作结束之后节点的位置;

【题解】



这个节点的标志方式类似树状数组;

用树状数组左走右走就好;

L->x-=lowbit(x)/2;

R->x+=lowbit(x)/2;

U->先假设x是左儿子,然后依照规则求出父亲F,然后看看F的左儿子是不是真的是x,是的话爸爸就是F,否则x是右儿子,然后根据x是右儿子求出对应的父亲节点就好;



【Number Of WA】



0



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define ps push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%lld",&x)
#define ref(x) scanf("%lf",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 1e5+100;

LL n,q,x;
int len;
char s[N];

LL lowbit(LL x){return x&(-x);}

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rel(n),rel(q);
    while (q--)
    {
        rel(x);
        scanf("%s",s+1);
        len = strlen(s+1);
        rep1(i,1,len)
        {
            LL t = lowbit(x);
            if (s[i]=='L')//向左走
            {
                x-=t/2;
            }
            else
                if (s[i]=='R')//向右走
                {
                    x+=t/2;
                }
                else
                {
                    if (x==(n+1)/2) continue;//是根节点就跳过
                    //假设x是爸爸的左儿子
                    LL baba = x+t;
                    LL zbaba = baba-lowbit(baba)/2;//然后求出这个假装爸爸的真左儿子
                    if (zbaba==x)
                    {
                        x = baba;
                    }
                    else//是右儿子;
                    {
                        x = x-t;
                    }
                }
        }
        printf("%lld\n",x);
    }
    //printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);
    return 0;
}