LCA Codeforces 100685G Gadget Hackwrench

 

题目传送门

题意：一棵有向的树，问u到v是否可达

分析：假设是无向树，DFS时正向的权值+1，反向的权值-1，然后找到LCA后判断dep数组和d数组就可以了

 

/************************************************
* Author        :Running_Time
* Created Time  :2015/10/5 星期一 10:28:49
* File Name     :G_2.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int D = 20;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
struct Edge {
    int v, d, nex;
}edge[N<<1];
int rt[D][N];
int dep[N], d[N];
int head[N], e;

void init(void) {
    memset (head, -1, sizeof (head));
    e = 0;
}

void add_edge(int u, int v, int dir) {
    edge[e] = (Edge) {v, dir, head[u]};
    head[u] = e++;
}

void DFS(int u, int fa, int dis)  {
    dep[u] = dep[fa] + 1;
    d[u] = dis;
    for (int i=head[u]; ~i; i=edge[i].nex)  {
        int v = edge[i].v;
        if (v == fa)    continue;
        rt[0][v] = u;
        DFS (v, u, dis + edge[i].d);
    }
}

int LCA(int u, int v)  {
    if (dep[u] < dep[v])    {
        swap (u, v);
    }
    for (int i=0; i<D; ++i)  {
        if ((dep[u] - dep[v]) >> i & 1)    {
            u = rt[i][u];
        }
    }
    if (u == v) return u;
    for (int i=D-1; i>=0; --i)  {
        if (rt[i][u] != rt[i][v])   {
            u = rt[i][u];
            v = rt[i][v];
        }
    }
    return rt[0][v];
}

int main(void)    {
    int n;
    while (scanf ("%d", &n) == 1)   {
        init ();
        for (int u, v, i=1; i<n; ++i) {
            scanf ("%d%d", &u, &v);
            add_edge (u, v, 1);
            add_edge (v, u, -1);
        }
        DFS (1, 0, 0);
        for (int i=1; i<D; ++i) {
            for (int j=1; j<=n; ++j)    {
                rt[i][j] = rt[i-1][rt[i-1][j]];
            }
        }
        int m;  scanf ("%d", &m);
        while (m--) {
            int u, v;   scanf ("%d%d", &u, &v);
            int f = LCA (u, v);
            if (dep[u] - dep[f] != d[f] - d[u]) puts ("No");
            else if (dep[v] - dep[f] != d[v] - d[f])    puts ("No");
            else    puts ("Yes");
        }
    }

    return 0;
}