标签：倍增 ext back query sort uniq def type 题解
题意：n个点的树，每个点有权值，问你u~v路径第k小的点的权值是？
思路：
树上主席树就是每个点建一棵权值线段树，具体看JQ博客，LCA用倍增logn求出，具体原理看这里
树上主席树我每个点的存的是点u到源点1的权值线段树，那我求点u到v的所有点，显然是 u + v - lca - fa[lca]，就是u到1 + v到1 - 多算的lca - 多算的fa[lca]。不能减去两个lca不然少一个点了，
LCA板子：
//LCA int fa[maxn][20]; int dep[maxn]; void lca_dfs(int u, int pre, int d){ dep[u] = d; fa[u][0] = pre; for(int i = head[u]; i != -1; i = edge[i].next) if(edge[i].v != pre) lca_dfs(edge[i].v, u, d + 1); } void lca_update(){ for (int i = 1; (1 << i) <= n; i++) for(int u = 1; u <= n; u++) fa[u][i] = fa[fa[u][i - 1]][i - 1]; } int lca_query(int u, int v){ if(dep[u] < dep[v]) swap(u, v); int d = dep[u] - dep[v]; for(int i = 0; (1 << i) <= d; i++) { if(d & (1 << i)) { u = fa[u][i]; } } if(u != v) { for(int i = (int)log2(n); i >= 0; i--) { if(fa[u][i] != fa[v][i]) { u = fa[u][i]; v = fa[v][i]; } } u = fa[u][0]; } return u; }
代码：
#include<cmath> #include<set> #include<queue> #include<cstdio> #include<vector> #include<cstring> #include <iostream> #include<algorithm> using namespace std; typedef long long ll; typedef unsigned long long ull; const int maxn = 100000 + 10; const int M = maxn * 30; const ull seed = 131; const int INF = 0x3f3f3f3f; const int MOD = 1000000007; int n, m; int root[maxn], a[maxn], tot; struct Edge{ int v, next; }edge[maxn << 1]; int head[maxn], tol; void addEdge(int u, int v){ edge[tol].v = v; edge[tol].next = head[u]; head[u] = tol++; } struct node{ int lson, rson; int sum; }T[maxn * 40]; void init(){ memset(T, 0, sizeof(T)); memset(root ,0, sizeof(root)); memset(head, -1, sizeof(head)); tot = tol = 0; } vector<int> ve; int getid(int x){ return lower_bound(ve.begin(), ve.end(), x) - ve.begin() + 1; } void update(int l, int r, int &now, int pre, int v, int pos){ T[++tot] = T[pre], T[tot].sum += v, now = tot; if(l == r) return; int m = (l + r) >> 1; if(pos <= m) update(l, m, T[now].lson, T[pre].lson, v, pos); else update(m + 1, r, T[now].rson, T[pre].rson, v, pos); } void build(int now, int pre){ update(1, n, root[now], root[pre], 1, getid(a[now])); for(int i = head[now]; i != -1; i = edge[i].next){ int v = edge[i].v; if(v == pre) continue; build(v, now); } } int query(int l, int r, int now, int pre, int lca, int flca, int k){ if(l == r) return l; int m = (l + r) >> 1; int sum = T[T[now].lson].sum + T[T[pre].lson].sum - T[T[lca].lson].sum - T[T[flca].lson].sum; if(sum >= k) return query(l, m, T[now].lson, T[pre].lson, T[lca].lson, T[flca].lson, k); else return query(m + 1, r, T[now].rson, T[pre].rson, T[lca].rson, T[flca].rson, k - sum); } //LCA int fa[maxn][20]; int dep[maxn]; void lca_dfs(int u, int pre, int d){ dep[u] = d; fa[u][0] = pre; for(int i = head[u]; i != -1; i = edge[i].next) if(edge[i].v != pre) lca_dfs(edge[i].v, u, d + 1); } void lca_update(){ for (int i = 1; (1 << i) <= n; i++) for(int u = 1; u <= n; u++) fa[u][i] = fa[fa[u][i - 1]][i - 1]; } int lca_query(int u, int v){ if(dep[u] < dep[v]) swap(u, v); int d = dep[u] - dep[v]; for(int i = 0; (1 << i) <= d; i++) { if(d & (1 << i)) { u = fa[u][i]; } } if(u != v) { for(int i = (int)log2(n); i >= 0; i--) { if(fa[u][i] != fa[v][i]) { u = fa[u][i]; v = fa[v][i]; } } u = fa[u][0]; } return u; } int main(){ init(); ve.clear(); scanf("%d%d", &n, &m); for(int i = 1; i <= n; i++) scanf("%d", &a[i]), ve.push_back(a[i]); sort(ve.begin(), ve.end()); ve.erase(unique(ve.begin(), ve.end()), ve.end()); for(int i = 1; i <= n - 1; i++){ int u, v; scanf("%d%d", &u, &v); addEdge(u, v); addEdge(v, u); } lca_dfs(1, 0, 1); lca_update(); build(1, 0); while(m--){ int u, v, k; scanf("%d%d%d", &u, &v, &k); int lca = lca_query(u, v); int ans = query(1, n, root[u], root[v], root[lca], root[fa[lca][0]], k); printf("%d\n", ve[ans - 1]); } return 0; }
SPOJ COT Count on a tree（树上主席树 + LCA 求路径第k小）题解
标签：倍增 ext back query sort uniq def type 题解
原文地址：https://www.cnblogs.com/KirinSB/p/10759869.html
