SPOJ COT Count on a tree 树上第k大（主席树）

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<
         for(int j = 1; j <= n - (1<
             dp[i][j] = dep[dp[i-1][j]] < dep[dp[i-1][j+(1<<(i-1))]] ? dp[i-1][j] : dp[i-1][j+(1<<(i-1))];
 }
 int RMQ(int l, int r)
 {
     int k = log(r - l + 1) / log(2);
     return dep[dp[k][l]] < dep[dp[k][r-(1<
 }
 int LCA(int v, int u)
 {
     v = fir[v], u = fir[u];
     if(v > 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;
 }