LCA转RMQ

来源：互联网发布：windows 系统错误编辑：程序博客网时间：2024/05/17 02:53

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;}

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