题意:http://www.spoj.com/problems/COT/en/
题意:给定一棵树,树上每个节点都有一个权值,问两点之间路径上第K大值
思路:树上的第k大值,跟区间第k大有些不同,区间第k大每个值在前一个值的基础上新建一棵树,而树上第k大则是在父亲节点的基础上新建一棵树。查询的时候,答案就是root[v] + root[u] - root[lca(v, u)] - root[fa[lca(v,u)]]上的第k大
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int N = 100010;
struct edge
{
int to, next;
}g[N*2];
int n, m, tot;
int cnt, head[N];
int len, root[N], lson[N*20], rson[N*20], val[N*20];
int num, dep[N*2], ver[N*2], fir[N], dp[20][N*2], fat[N];
bool vis[N];
int a[N], b[N];
void add_edge(int v, int u)
{
g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;
}
void build(int l, int r, int &rt)
{
rt = ++tot;
val[rt] = 0;
if(l == r) return;
int mid = (l + r) >> 1;
build(l, mid, lson[rt]);
build(mid + 1, r, rson[rt]);
}
void update(int pre, int &rt, int l, int r, int v)
{
rt = ++tot;
lson[rt] = lson[pre], rson[rt] = rson[pre], val[rt] = val[pre] + 1;
if(l == r) return;
int mid = (l + r) >> 1;
if(v <= mid) update(lson[pre], lson[rt], l, mid, v);
else update(rson[pre], rson[rt], mid + 1, r, v);
}
void dfs(int v, int fa, int d)
{
vis[v] = true, ver[++num] = v, dep[num] = d, fir[v] = num, fat[v] = fa;
update(root[fa], root[v], 1, len, a[v]);//在父节点的基础上新建一棵树
for(int i = head[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(! vis[u])
{
dfs(u, v, d + 1);
ver[++num] = v, dep[num] = d;
}
}
}
void ST(int n)
{
for(int i = 1; i <= n; i++) dp[0][i] = i;
for(int i = 1; (1< u) swap(v, u);
int res = RMQ(v, u);
return ver[res];
}
int query(int ss, int tt, int lca, int lca_fa, int l, int r, int k)
{
if(l == r) return l;
int mid = (l + r) >> 1;
int tmp = val[lson[ss]] + val[lson[tt]] - val[lson[lca]] - val[lson[lca_fa]];
if(k <= tmp) return query(lson[ss], lson[tt], lson[lca], lson[lca_fa], l, mid, k);
else return query(rson[ss], rson[tt], rson[lca], rson[lca_fa], mid + 1, r, k - tmp);
}
int main()
{
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
sort(b+1, b+1+n);
len = unique(b+1, b+1+n) - b - 1;
for(int i = 1; i <= n; i++) a[i] = lower_bound(b+1, b+1+len, a[i]) - b;
cnt = 0;
memset(head, -1, sizeof head);
for(int i = 1; i <= n - 1; i++)
{
int v, u;
scanf("%d%d", &v, &u);
add_edge(v, u), add_edge(u, v);
}
tot = 0;
build(1, len, root[0]);
num = 0;
memset(vis, 0, sizeof vis);
dfs(1, 0, 1);
ST(2 * n - 1);
for(int i = 1; i <= m; i++)
{
int v, u, k;
scanf("%d%d%d", &v, &u, &k);
int lca = LCA(v, u);
printf("%d\n", b[query(root[v], root[u], root[lca], root[fat[lca]], 1, len , k)]);
}
return 0;
}