SPOJ

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 (u, v).

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 5105 2 9 3 8 5 7 71 21 31 43 53 63 74 82 5 12 5 22 5 32 5 47 8 2 

Output:2891057 

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)]。

类似的，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];//LCAint 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_Treeint 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 ProLights

  Thank you!