LCA转RMQ

不知道有没有打错，先保存一下

#include <iostream>
#include <cstring>
#include <cstdio>
#Inlcude <queue>
#include <algorithm>
using namespace std;
const int N = 10000;
int dfs_clock,in[N],out[N],dep[N],pos[N]dp[N][20];
void init()
{
    dfs_clock = 0;
}

void dfs(int cur,int d)
{
    in[cur] = ++dfs_clock;
    pos[dfs_clock] = cur;
    dep[dfs_clock] = d;
    for(int i = eh[cur],i != -1; i = edge[i].next){
            dfs(edge[i].v,d + 1);
    }
    out[cur] = ++dfs_clock;
    pos[dfs_clock] = cur;
    dep[dfs_clock] = d;
}
void RMQ_init()//保存深度最小的值的下标
{   
    for(int i = 1; i <= dfs_clock; ++i) dp[i][0] = i;
    int len = int(log(dfs_clock)/log(2));
    for(int j = 1; j <= len; ++j){
        for(int i = 1; i <= dfs_clock ; ++i){
                if(dep[dp[i][j - 1]] < dep[dp[i + (1 <<(j - 1))][j - 1]])
                        dp[i][j] = dp[i][j - 1];
                else dp[i][j] = dp[i + 1 << (j - 1)][j - 1];
        }
    }
}
int RMQ(int L,int R)//返回最近公共祖先在原数组的下标
{
    int k = int(log(R - L + 1)/log(2));
    if(dep[dp[L][k]] < dep[dp[R - (1 << k) + 1][k]])
            return pos[dp[L][K]];
    else return pos[dp[R - (1 << k) + 1][k]];
}

int mian()
{
    
    
    
    
    return 0;
}