BZOJ 3910 火车 LCA+并查集

题目大意

给出一棵树，起点，和要经过的点的序列，已经经过的点就不用去了，剩下的点按照顺序依次去，问要经过多少条边。

思路

链剖大概应该是可以，不过没试，用了听大爷说的一种神奇的方法。
因为树上经过的点肯定是一段一段的，就想到用并查集将一段合成一个点，每个点最多只能被合一次，这样的话就能保证时间复杂度。查询的时候像链剖一样一段一段往上跳就行了，还要顺便把路径上的所有点缩起来。

CODE

#define _CRT_SECURE_NO_WARNINGS

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 500010
using namespace std;

int points, cross, start;
int head[MAX], total;
int _next[MAX << 1], aim[MAX << 1];

inline void Add(int x, int y)
{
    _next[++total] = head[x];
    aim[total] = y;
    head[x] = total;
}

int deep[MAX];
int father[MAX][20];

void DFS(int x, int last)
{
    deep[x] = deep[last] + 1;
    for(int i = head[x]; i; i = _next[i]) {
        if(aim[i] == last)  continue;
        father[aim[i]][0] = x;
        DFS(aim[i], x);
    }
}

void SparseTable()
{
    for(int j = 1; j <= 19; ++j)
        for(int i = 1; i <= points; ++i)
            father[i][j] = father[father[i][j - 1]][j - 1];
}

inline pair<int, int> GetLCA(int x, int y)
{
    pair<int, int> re;
    if(deep[x] < deep[y])   swap(x, y);
    for(int i = 19; ~i; --i)
        if(deep[father[x][i]] >= deep[y]) {
            re.second += 1 << i;
            x = father[x][i];
        }
    if(x == y) {
        re.first = x;
        return re;
    }
    for(int i = 19; ~i; --i)
        if(father[x][i] != father[y][i]) {
            re.second += 1 << (i + 1);
            x = father[x][i];
            y = father[y][i];
        }
    re.first = father[x][0];
    re.second += 2;
    return re;
}

namespace UnionFindSet{
    int father[MAX];

    int Find(int x) {
        if(father[x] == x)  return x;
        return father[x] = Find(father[x]);
    }
}

inline void Work(int x, int y, int lca)
{
    using namespace UnionFindSet;
    x = Find(x), y = Find(y);
    while(x != y) {
        if(deep[Find(::father[x][0])] < deep[Find(::father[y][0])]) swap(x, y);
        UnionFindSet::father[x] = ::father[x][0];
        x = Find(x);
    }
    if(x == lca)
        UnionFindSet::father[lca] = ::father[lca][0] ? ::father[lca][0]:points + 1;
}

int main()
{
    cin >> points >> cross >> start;
    for(int x, y, i = 1; i < points; ++i) {
        scanf("%d%d", &x, &y);
        Add(x, y), Add(y, x);
    }
    DFS(1, 0);
    SparseTable();
    for(int i = 1; i <= points; ++i)
        UnionFindSet::father[i] = i;
    long long ans = 0;
    int now = start;
    for(int x, i = 1; i <= cross; ++i) {
        scanf("%d", &x);
        int fx = UnionFindSet::Find(x);
        if(fx != x) continue;
        pair<int ,int> re = GetLCA(x, now);
        ans += re.second;
        Work(x, now, re.first);
        now = x;
    }
    cout << ans << endl;
    return 0;
}