SPOJ - COT Count on a tree 树上主席树+LCA+任意路径问题
Count on a tree
SPOJ - COTYou 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 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)]。
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
#include
#include
#include
using namespace std;
typedef long long LL;
const int MAXN = 1e5 + 8;
vector 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<= 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 ProLights
Thank you!