Orz..跑得还挺快的#10
自从会树链剖分后LCA就没写过倍增了...
这道题用可持久化线段树..点x的线段树表示ROOT到x的这条路径上的权值线段树
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#include<bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < n; i++)
#define clr(x, c) memset(x, c, sizeof(x))
#define M(l, r) (((l) + (r)) >> 1)
#define foreach(i, x) for(__typeof(x.begin()) i = x.begin(); i != x.end(); i++)
using namespace std;
const int maxn = 200009;
int w[maxn], n, N, id[maxn];
vector<int> G[maxn];
namespace LCA {
int top[maxn], son[maxn], dep[maxn], size[maxn], fa[maxn], TOP;
void dfs(int x) {
size[x] = 1; son[x] = -1;
foreach(it, G[x]) if(*it != fa[x]) {
dep[*it] = dep[x] + 1;
fa[*it] = x;
dfs(*it);
size[x] += size[*it];
if(!~son[x] || size[*it] > size[son[x]])
son[x] = *it;
}
}
void DFS(int x) {
top[x] = TOP;
if(~son[x]) DFS(son[x]);
foreach(it, G[x]) if(*it != son[x] && *it != fa[x])
DFS(TOP = *it);
}
void init() {
dep[0] = 0; fa[0] = -1;
dfs(0); DFS(TOP = 0);
}
int LCA(int x, int y) {
for(; top[x] != top[y]; x = fa[top[x]])
if(dep[top[x]] < dep[top[y]]) swap(x, y);
return dep[x] < dep[y] ? x : y;
}
}
struct Node {
Node *l, *r;
int s;
} pool[maxn * 40], *pt = pool, *null, *root[maxn];
void init() {
null = pt++; null->s = 0;
null->l = null->r = null;
}
int v;
Node* modify(Node* t, int l, int r) {
Node* h = pt++;
h->s = t->s + 1;
if(r > l) {
int m = M(l, r);
if(v <= m) {
h->l = modify(t->l, l, m);
h->r = t->r;
} else {
h->l = t->l;
h->r = modify(t->r, m + 1, r);
}
}
return h;
}
void build(int x, int fa = -1, Node* p = null) {
v = w[x] + 1;
root[x] = modify(p, 1, N);
foreach(it, G[x]) if(*it != fa)
build(*it, x, root[x]);
}
inline int read() {
char c = getchar();
int ans = 0, f = 1;
for(; !isdigit(c); c = getchar())
if(c == '-') f = -1;
for(; isdigit(c); c = getchar())
ans = ans * 10 + c - '0';
return ans * f;
}
int ans = 0;
void work(bool F) {
int x = (read()^ans) - 1, y = read() - 1, k = read(), lca = LCA::LCA(x, y);
Node *X = root[x], *Y = root[y], *A = root[lca], *P = lca ? root[LCA::fa[lca]] : null;
int L = 1, R = N;
while(L < R) {
int s = X->l->s + Y->l->s - A->l->s - P->l->s, m = M(L, R);
if(s >= k) {
X = X->l; Y = Y->l; A = A->l; P = P->l;
R = m;
} else {
X = X->r; Y = Y->r; A = A->r; P = P->r;
k -= s;
L = m + 1;
}
}
ans = id[L - 1];
printf("%d", ans);
if(F) putchar('\n');
}
int main() {
freopen("test.in", "r", stdin);
init();
n = read();
int m = read();
rep(i, n) {
id[i] = w[i] = read();
G[i].clear();
}
sort(id, id + n);
N = unique(id, id + n) - id;
rep(i, n) w[i] = lower_bound(id, id + N, w[i]) - id;
rep(i, n - 1) {
int u = read() - 1, v = read() - 1;
G[u].push_back(v);
G[v].push_back(u);
}
LCA::init();
build(0);
while(m--) work(m);
return 0;
}
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2588: Spoj 10628. Count on a tree
Time Limit: 12 Sec Memory Limit: 128 MB
Submit: 2675 Solved: 606
[Submit][Status][Discuss]
Description
给定一棵N个节点的树，每个点有一个权值，对于M个询问(u,v,k)，你需要回答u xor lastans和v这两个节点间第K小的点权。其中lastans是上一个询问的答案，初始为0，即第一个询问的u是明文。
Input
第一行两个整数N,M。
第二行有N个整数，其中第i个整数表示点i的权值。
后面N-1行每行两个整数(x,y)，表示点x到点y有一条边。
最后M行每行两个整数(u,v,k)，表示一组询问。
Output
M行，表示每个询问的答案。
Sample 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
0 5 2
10 5 3
11 5 4
110 8 2
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
0 5 2
10 5 3
11 5 4
110 8 2
Sample Output
2
8
9
105
7
8
9
105
7
HINT
HINT：
N,M<=100000
暴力自重。。。