SPOJ COT Count on a tree 主席树

题意：

给\(n(1 \leq n \leq 10^5)\)一棵树，每个点有个权值。

还有\(m(1 \leq m \leq 10^5)\)个询问：

\(u \, v \, k\)，查询路径\(u \to v\)上权值第\(k\)小的权值。

分析：

和普通的区间一样，对于树维护到根节点的路径信息，父亲节点代表的树就是它的前一棵树。

查询的时候还要求出\(LCA(u,v)\)，路径上的点数就是：

\(u\)到根的个数+\(v\)到根的个数-\(lca\)到根的个数\(\times 2\)

如果\(lca\)的权值也在统计范围内，统计个数再加\(1\)。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 100000 + 10;

struct Edge
{
int v, nxt;
Edge() {}
Edge(int v, int nxt): v(v), nxt(nxt) {}
};

int head[maxn], ecnt;
Edge edges[maxn * 2];

void AddEdge(int u, int v) {
edges[ecnt] = Edge(v, head[u]);
head[u] = ecnt++;
}

int fa[maxn], L[maxn];

void dfs(int u) {
for(int i = head[u]; ~i; i = edges[i].nxt) {
int v = edges[i].v;
if(v == fa[u]) continue;
fa[v] = u;
L[v] = L[u] + 1;
dfs(v);
}
}

int n, m;

int anc[maxn][20];

void preprocess() {
memset(anc, 0, sizeof(anc));
for(int i = 1; i <= n; i++) anc[i][0] = fa[i];
for(int j = 1; (1 << j) < n; j++)
for(int i = 1; i <= n; i++) if(anc[i][j-1])
anc[i][j] = anc[anc[i][j-1]][j-1];
}

int LCA(int u, int v) {
if(L[u] < L[v]) swap(u, v);
int log;
for(log = 0; (1 << log) < L[u]; log++);
for(int i = log; i >= 0; i--)
if(L[u] - (1<<i) >= L[v]) u = anc[u][i];
if(u == v) return u;
for(int i = log; i >= 0; i--)
if(anc[u][i] && anc[u][i] != anc[v][i])
u = anc[u][i], v = anc[v][i];
return fa[u];
}

int a[maxn], b[maxn], tot;

struct Node
{
int lch, rch, sum;
};

int sz;
Node T[maxn << 5];
int root[maxn];

int newnode() {
sz++;
T[sz].lch = T[sz].rch = T[sz].sum = 0;
return sz;
}

int update(int pre, int L, int R, int p) {
int rt = newnode();
T[rt].lch = T[pre].lch;
T[rt].rch = T[pre].rch;
T[rt].sum = T[pre].sum + 1;
if(L == R) return rt;
int M = (L + R) / 2;
if(p <= M) T[rt].lch = update(T[pre].lch, L, M, p);
else T[rt].rch = update(T[pre].rch, M+1, R, p);
return rt;
}

void dfs2(int u) {
root[u] = update(root[fa[u]], 1, tot, a[u]);
for(int i = head[u]; ~i; i = edges[i].nxt) {
int v = edges[i].v;
if(v == fa[u]) continue;
dfs2(v);
}
}

int tmp;

int query(int u, int v, int l, int L, int R, int k) {
if(L == R) return L;
int M = (L + R) / 2;
int sum = T[T[u].lch].sum + T[T[v].lch].sum - T[T[l].lch].sum * 2;
if(L <= tmp && tmp <= M) sum++;
if(sum >= k) return query(T[u].lch, T[v].lch, T[l].lch, L, M, k);
else return query(T[u].rch, T[v].rch, T[l].rch, M+1, R, k - sum);
}

int main()
{
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++) {
scanf("%d", a + i);
b[i] = a[i];
}
ecnt = 0;
memset(head, -1, sizeof(head));
for(int i = 1; i < n; i++) {
int u, v; scanf("%d%d", &u, &v);
AddEdge(u, v); AddEdge(v, u);
}

dfs(1);
preprocess();

sort(b + 1, b + 1 + n);
tot = unique(b + 1, b + 1 + n) - b - 1;
for(int i = 1; i <= n; i++)
a[i] = lower_bound(b + 1, b + 1 + tot, a[i]) - b;
dfs2(1);

while(m--) {
int u, v, k; scanf("%d%d%d", &u, &v, &k);
int lca = LCA(u, v);
tmp = a[lca];
int ans = query(root[u], root[v], root[lca], 1, tot, k);
printf("%d\n", b[ans]);
}

return 0;
}