2015北京网络赛 Couple Trees 倍增算法

2015北京网络赛 Couple Trees

题意：两棵树，求不同树上两个节点的最近公共祖先

思路：比赛时看过的队伍不是很多，没有仔细想。今天补题才发现有个 倍增算法，自己竟然不知道。

　　  解法来自 qscqesze ，这个其实之前如果了解过倍增的话还是不是很难，不过这题的数据也不是很给力，极限数据理论上是过不了的。

　　  其他解法有树链剖分？并不是很清楚。就这样水过了吧。。。

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define LL long long
#define eps 1e-8
#define INF 0x3f3f3f3f
#define MAXN 100005
using namespace std;

int f1[MAXN][20], f2[MAXN][20];
int deep1[MAXN], deep2[MAXN];
int step1, step2, ans;
void work(int x, int y){
    step1 = step2 = 1;
    while (x != y){
        if (x < y){
            //x < y means y is not x's ancestor, so let y up
            for (int i = 15; i >= 0; i--){
                if (f2[y][i] > x){
                    y = f2[y][i];
                    step2 += 1 << i;
                    break;
                }
            }
            y = f2[y][0];
            step2++;
        }
        else{
            for (int i = 15; i >= 0; i--){
                if (f1[x][i] > y){
                    x = f1[x][i];
                    step1 += 1 << i;
                    break;
                }
            }
            x = f1[x][0];
            step1++;
        }
    }
    ans = x;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // OPEN_FILE
    int n, m;
    while (~scanf("%d%d", &n, &m)){
        int x, y;
        deep1[1] = deep2[1] = 1;
        for (int i = 0; i <= 15; i++){
            f1[1][i] = f2[1][i] = 1;
        }
        for (int i = 2; i <= n; i++){
            scanf("%d", &x);
            f1[i][0] = x;
            deep1[i] = deep1[x] + 1;
            for (int j = 1; j <= 15; j++){
                f1[i][j] = f1[f1[i][j - 1]][j - 1];
            }
        }
        for (int i = 2; i <= n; i++){
            scanf("%d", &x);
            f2[i][0] = x;
            deep2[i] = deep2[x] + 1;
            for (int j = 1; j <= 15; j++){
                f2[i][j] = f2[f2[i][j - 1]][j - 1];
            }
        }
        ans = 0;
        for (int i = 1; i <= m; i++){
            scanf("%d%d", &x, &y);
            x = (x + ans) % n + 1;
            y = (y + ans) % n + 1;
            work(x, y);
            printf("%d %d %d
", ans, step1, step2);
        }
    }
}