SPOJ – COT Count on a tree 树上主席树+LCA+任意路径问题

云计算 waitig 558℃ 百度已收录 0评论

Count on a tree

 SPOJ
– COT

You are given a tree with N nodes. The tree nodes are numbered from 1 to N. Each node has an integer weight.

We will ask you to perform the following operation:

  • u v k : ask for the kth minimum weight on the path from node u to node v

Input

In the first line there are two integers N and M. (N, M <= 100000)

In the second line there are N integers. The ith integer denotes the weight of the ith node.

In the next N-1 lines, each line contains two integers u v, which describes an edge (uv).

In the next M lines, each line contains three integers u v k, which means an operation asking for the kth minimum weight on the path from node u to
node v.

Output

For each operation, print its result.

Example

Input:
8 5
105 2 9 3 8 5 7 7
1 2
1 3
1 4
3 5
3 6
3 7
4 8
2 5 1
2 5 2
2 5 3
2 5 4
7 8 2 
Output:
2
8
9
105
7 

Source:
SPOJ - COT 
[kuangbin]主席树]

My Solution:
题意:给出一个树和树上每个点的权值,给出m个询问(u,v,k),询问在树上从点u到点v所构成的路径上权值第k小的点的权值。

树上主席树+LCA+任意路径问题
主席树维护的一个前缀和,而前缀和不一定要出现在一个线性表上(或者说树型可以线性话),即对于每个从根到点u的路径是一个线性序列,可以把这个序列建成主席树。
利用这个前缀和,我们可以解决一些树上任意路径的问题,比如在线询问[u,v]点对的距离——dist[u]+dist[v]-2*dist[LCA(u,v)]。
同理,我们可以利用主席树来解决树上任意路径的问题。
DFS遍历整棵树,然后在每个节点上建立一棵线段树,某一棵线段树的“前一版本”是位于该节点父亲节点fa的线段树。那么对于询问[a,b],答案就是root[u]+root[v]-root[LCA(u,v)]-root[fa[LCA(u,v)]]上的第k小。
时间复杂度 O(nlogn)
空间复杂度 O(nlogn)
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MAXN = 1e5 + 8;

vector<int> sons[MAXN];

//LCA
int father[MAXN][22], depth[MAXN];
//!LCA 倍增算法 可以动态加减叶子节点, 因为预处理的时候每个叶子是独立的
inline void prepare(int n){
    int i, j;
    for(j = 1; j <= 20; j++){
        for(i = 1; i <= n; i++){
            father[i][j] = father[father[i][j-1]][j-1];
        }
    }
}
inline int LCA(int u, int v) {
    if (depth[u] < depth[v]) swap(u, v);
    int dc = depth[u] - depth[v];
    int i;
    for(i = 0; i <= 20; i++){
        if((1<<i) & dc) u = father[u][i];
    }
    if(u == v) return u;
    for(i = 20; i >= 0; i--){
        if (father[u][i] != father[v][i]){
            u = father[u][i];
            v = father[v][i];
        }
    }
    u = father[u][0];
    return u;
}


//Chairman_Tree
int tot, size;
int a[MAXN], b[MAXN], root[20*MAXN], ls[20*MAXN], rs[20*MAXN], sum[20*MAXN];

inline void _build(int &o, int l, int r){
    o = ++ tot;
    sum[o] = 0;
    if(l == r) return;
    int mid = (l + r) >> 1;
    _build(ls[o], l, mid);
    _build(rs[o], mid + 1, r);
}

inline void update(int &o, int l, int r, int last, int p){
    o = ++ tot;
    ls[o] = ls[last];
    rs[o] = rs[last];
    sum[o] = sum[last] + 1;
    if(l == r) return;
    int mid = (l + r) >> 1;
    if(p <= mid) update(ls[o], l, mid, ls[last], p);
    else update(rs[o], mid + 1, r, rs[last], p);
}

inline int _query(int u, int v, int lca, int flca, int l, int r, int k){
    if(l == r) return l;
    int mid = (l + r) >> 1;
    int cnt = sum[ls[u]] + sum[ls[v]] - sum[ls[lca]] - sum[ls[flca]];
    if(k <= cnt) return _query(ls[u], ls[v], ls[lca], ls[flca], l, mid, k);
    else return _query(rs[u], rs[v], rs[lca], rs[flca], mid + 1, r, k - cnt);
}

//
inline void dfs(int u, int fa){//
    int sz = sons[u].size(), i, v;
    //cout << u <<" "<< a[u] << endl;
    update(root[u], 1, size, root[fa], a[u]);
    for(i = 0; i < sz; i++){
        v = sons[u][i];
        if(v == fa) continue;
        father[v][0] = u;
        depth[v] = depth[u] + 1;
        dfs(v, u);
    }
}

int main()
{
    #ifdef LOCAL
    freopen("c.txt", "r", stdin);
    //freopen("c.out", "w", stdout);
    int T = 1;
    while(T--){
    #endif // LOCAL
    ios::sync_with_stdio(false); cin.tie(0);

    int n, m, i, u, v, x, ind, lca;
    cin >> n >> m;
    for(i = 1; i <= n; i++){
        cin >> a[i];
        b[i] = a[i];
    }
    for(i = 1; i < n; i++){
        cin >> u >> v;
        sons[u].push_back(v);
        sons[v].push_back(u);
    }
    sort(b + 1, b + n + 1);
    size = unique(b + 1, b + n + 1) - (b + 1);
    for(i = 1; i <= n; i++){
        //cout << a[i] << " ";
        a[i] = lower_bound(b + 1, b + size + 1, a[i]) - b; //It is based on 1~n..
        //cout << a[i] << "\n";
    }
    tot = 0;
    _build(root[0], 1, size);
    dfs(1, 0);
    prepare(n);
    while(m--){
        cin >> u >> v >> x;
        lca = LCA(u, v);
        //cout << u << " " << v << " " << lca << endl;
        ind = _query(root[u], root[v], root[lca], root[father[lca][0]], 1, size, x);
        cout << b[ind] << "\n";
    }


    #ifdef LOCAL
    cout << endl;
    }
    #endif // LOCAL
    return 0;
}

                                                                                                           
                                 ——from 
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  Thank you!


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