題意：

給\(n(1 \leq n \leq 10^5)\)一棵樹，每個點有個權(quán)值。
還有\(m(1 \leq m \leq 10^5)\)個詢問：
\(u \, v \, k\)，查詢路徑\(u \to v\)上權(quán)值第\(k\)小的權(quán)值。

分析：

和普通的區(qū)間一樣，對于樹維護到根節(jié)點的路徑信息，父親節(jié)點代表的樹就是它的前一棵樹。
查詢的時候還要求出\(LCA(u,v)\)，路徑上的點數(shù)就是：

• \(u\)到根的個數(shù)+\(v\)到根的個數(shù)-\(lca\)到根的個數(shù)\(\times 2\)

如果\(lca\)的權(quán)值也在統(tǒng)計范圍內(nèi)，統(tǒng)計個數(shù)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; }

轉(zhuǎn)載于:https://www.cnblogs.com/AOQNRMGYXLMV/p/5285890.html

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