[SPOJ 10628]Count on a tree
2018-02-28 15:37
295 查看
Description
题库链接求不带修改的树上路径第 \(K\) 小。 \(N\) 个节点 \(M\) 组询问。
\(1\leq N,M\leq 100000\)
Solution
主席树维护树上前缀和。分离一段路径的技巧: \(val_u+val_v-val_{lca_{u,v}}-val_{fa_{lca_{u,v}}}\) 。其余的就是 \(K-th~number\) 了。Code
//It is made by Awson on 2018.2.28 #include <bits/stdc++.h> #define LL long long #define dob complex<double> #define Abs(a) ((a) < 0 ? (-(a)) : (a)) #define Max(a, b) ((a) > (b) ? (a) : (b)) #define Min(a, b) ((a) < (b) ? (a) : (b)) #define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b)) #define writeln(x) (write(x), putchar('\n')) #define lowbit(x) ((x)&(-(x))) using namespace std; const int N = 100000; void read(int &x) { char ch; bool flag = 0; for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar()); for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar()); x *= 1-2*flag; } void print(int x) {if (x > 9) print(x/10); putchar(x%10+48); } void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); } int a[N+5], n, m, b[N+5], u, v, last, tot, k; struct tt {int to, next; }edge[(N<<1)+5]; int path[N+5], Top, dep[N+5], fa[N+5], top[N+5], size[N+5], son[N+5]; struct Segment_tree { int root[N+5], key[N*40+5], ch[N*40+5][2], pos; int cpynode(int o) {++pos, ch[pos][0] = ch[o][0], ch[pos][1] = ch[o][1], key[pos] = key[o]; return pos; } void insert(int &o, int l, int r, int loc) { o = cpynode(o); ++key[o]; if (l == r) return; int mid = (l+r)>>1; if (loc <= mid) insert(ch[o][0], l, mid, loc); else insert(ch[o][1], mid+1, r, loc); } int query(int a, int b, int c, int d, int l, int r, int k) { if (l == r) return l; int mid = (l+r)>>1; int tmp = key[ch[a][0]]+key[ch[b][0]]-key[ch[c][0]]-key[ch[d][0]]; if (tmp >= k) return query(ch[a][0], ch[b][0], ch[c][0], ch[d][0], l, mid, k); else return query(ch[a][1], ch[b][1], ch[c][1], ch[d][1], mid+1, r, k-tmp); } }T; void dfs1(int o, int father, int depth) { T.root[o] = T.root[father]; T.insert(T.root[o], 1, tot, lower_bound(b+1, b+tot+1, a[o])-b); size[o] = 1, dep[o] = depth, fa[o] = father; for (int i = path[o]; i; i = edge[i].next) if (edge[i].to != father) { dfs1(edge[i].to, o, depth+1); size[o] += size[edge[i].to]; if (size[edge[i].to] > size[son[o]]) son[o] = edge[i].to; } } void dfs2(int o, int tp) { top[o] = tp; if (son[o]) dfs2(son[o], tp); for (int i = path[o]; i; i = edge[i].next) if (edge[i].to != fa[o] && edge[i].to != son[o]) dfs2(edge[i].to, edge[i].to); } int get_lca(int x, int y) { while (top[x] != top[y]) { if (dep[top[x]] < dep[top[y]]) Swap(x, y); x = fa[top[x]]; } return dep[x] > dep[y] ? y : x; } void add(int u, int v) {edge[++Top].to = v, edge[Top].next = path[u], path[u] = Top; } void work() { read(n), read(m); for (int i = 1; i <= n; i++) read(a[i]), b[i] = a[i]; for (int i = 1; i < n; i++) read(u), read(v), add(u, v), add(v, u); sort(b+1, b+n+1); tot = unique(b+1, b+n+1)-b-1; dfs1(1, 0, 1); dfs2(1, 1); for (int i = 1; i <= m; i++) { read(u), read(v); u ^= last; read(k); int lca = get_lca(u, v); write(last = b[T.query(T.root[u], T.root[v], T.root[lca], T.root[fa[lca]], 1, tot, k)]); if (i != m) putchar('\n'); } } int main() { work(); return 0; }
相关文章推荐
- BZOJ 2588: Spoj 10628. Count on a tree [树上主席树]
- Spoj 10628. Count on a tree 题解
- 【BZOJ2588】【Spoj 10628.】 Count on a tree 可持久化线段树+lca
- SPOJ 10628 Count on The tree .. .. 主席树
- Spoj 10628. Count on a tree HYSBZ - 2588 (树链剖分+主席树)
- 2588: Spoj 10628. Count on a tree[可持久化线段树+倍增lca]
- 2588: Spoj 10628. Count on a tree 主席树+LCA
- bzoj2588: Spoj 10628. Count on a tree 主席树+dfs序
- BZOJ_2588_Spoj 10628. Count on a tree_树剖+主席树
- [bzoj2588][Spoj 10628] Count on a tree
- BZOJ 2588: Spoj 10628. Count on a tree
- BZOJ 2588: Spoj 10628. Count on a tree | 树上主席树
- BZOJ2588: Spoj 10628. Count on a tree
- BZOJ2588 Spoj 10628. Count on a tree
- 【bzoj2588】Spoj 10628. Count on a tree
- [主席树] BZOJ2588: Spoj 10628. Count on a tree
- SPOJ 10628 Count on a tree (lca+主席树)
- 2588: Spoj 10628. Count on a tree
- BZOJ2588: Spoj 10628. Count on a tree
- [主席树] BZOJ 2588 Spoj 10628. Count on a tree