bzoj2588: Spoj 10628. Count on a tree

题面在这里

题意：

给一棵树，多组询问，每次询问u,v路径上点权的第k小。

做法：

本题卡时间，只能带一个log。
考虑每个节点维护一个到根的权值线段树，查询x,y的路径上的和就是sum[x]+sum[y]-sum[lca(x,y)]-sum[fa[lca(x,y)]]。
于是每次只要把4个数传进参数里。qwq不用二分。

另外这题可以树链+二分+树套树，log^4，不过会T.

代码：

/*************************************************************
    Problem: bzoj 2588 Spoj 10628. Count on a tree
    User: fengyuan
    Language: C++
    Result: Accepted
    Time: 5836 ms
    Memory: 37932 kb
    Submit_Time: 2017-12-26 08:13:18
*************************************************************/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#define mid ((l+r)/2)
using namespace std;

const int N = 100010, M = 2000010;
int n, m, cnt, tot;
int head[N], s[N], b[N], rt[N], depth[N], f[N][19];
int L[M], R[M], sum[M];
struct Node{ int num, id; }a[N];
struct Edge{ int to, nex; }e[N<<1];

inline void addEdge(int x, int y)
{
    e[++ cnt].to = y;
    e[cnt].nex = head[x];
    head[x] = cnt;
}

const bool cmp(const Node &x, const Node &y){ return x.num < y.num; }

inline void insert(int pre, int &rt, int l, int r, int pos)
{
    rt = ++ tot; sum[rt] = sum[pre] + 1;
    if(l == r) { s[l] = pos; return; }
    if(pos <= mid) R[rt] = R[pre], insert(L[pre], L[rt], l, mid, pos);
    else L[rt] = L[pre], insert(R[pre], R[rt], mid+1, r, pos);
}

inline void dfs(int u, int last, int s)
{
    depth[u] = s; f[u][0] = last;
    insert(rt[last], rt[u], 1, n, b[u]);
    for(int i = head[u]; i; i = e[i].nex) {
        int v = e[i].to; if(v == last) continue;
        dfs(v, u, s + 1);
    }
}

inline int LCA(int x, int y)
{
    if(depth[x] < depth[y]) swap(x, y);
    int tmp = depth[x] - depth[y];
    for(int i = 17; i >= 0; i --)
        if((tmp>>i)&1) x = f[x][i];
    if(x == y) return x;
    for(int i = 17; i >= 0; i --)
        if(f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
    return f[x][0];
}

inline int query(int u, int v, int lca, int fa_lca, int l, int r, int K)
{
    if(l == r) return a[s[l]].num;
    int w = sum[L[u]] + sum[L[v]] - sum[L[lca]] - sum[L[fa_lca]];
    if(w >= K) return query(L[u], L[v], L[lca], L[fa_lca], l, mid, K);
    else return query(R[u], R[v], R[lca], R[fa_lca], mid+1, r, K-w);
}

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i ++) scanf("%d", &a[i].num), a[i].id = i;
    sort(a+1, a+1+n, cmp);
    for(int i = 1; i <= n; i ++) b[a[i].id] = i;
    for(int i = 1; i < n; i ++) {
        int x, y; scanf("%d%d", &x, &y);
        addEdge(x, y); addEdge(y, x);
    }
    dfs(1, 0, 0);
    for(int j = 1; j <= 17; j ++)
        for(int i = 1; i <= n; i ++) f[i][j] = f[f[i][j-1]][j-1];
    int last = 0;
    while(m --) {
        int u, v, K; scanf("%d%d%d", &u, &v, &K);
        u ^= last;
        int w = LCA(u, v);
        last = query(rt[u], rt[v], rt[w], rt[f[w][0]], 1, n, K);
        printf("%d", last); if(m) putchar('\n');
    }
    return 0;
}