HDU 3078 Network （lca+排序）

題意：一棵無向樹，輸入點數和操作數，下面一行n個值代表每個點的權。下面n-1行是樹邊。操作分為 0 x w ，表示把點x的權改為w； k a b ， 求出從a到b的路徑中，第k大的點權。

題解：lca+排序

居然直接暴力就可以過了，每次詢問将lca路徑中的點權排序即可。

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<fstream>
#include<set>
#include<map>
#include<sstream>
#include<iomanip>
#define ll long long
#define pii pair<int, int>
using namespace std;
const int MAXN = 8e4 + 5;
int n, q;
vector<int> v[MAXN];
int fa[MAXN][31], dep[MAXN];
void init() {
    for (int i = 1; i <= n; i++) v[i].clear();
    // memset(fa, 0, sizeof(fa));
    // memset(cost, 0, sizeof(cost));
    // memset(dep, 0, sizeof(dep));
}
void dfs(int root, int fno) {   //1 0
    fa[root][0] = fno;
    dep[root] = dep[fa[root][0]] + 1;
    for (int i = 1; i < 31; ++i) {
        fa[root][i] = fa[fa[root][i - 1]][i - 1];
    }
    int sz = v[root].size();
    for (int i = 0; i < sz; ++i) {
        if (v[root][i] == fno) continue;
        dfs(v[root][i], root);
    }
}
int lca(int x, int y) {
    if (dep[x] > dep[y]) swap(x, y);
    int tmp = dep[y] - dep[x];
    for (int j = 0; tmp; ++j, tmp >>= 1)
        if (tmp & 1) y = fa[y][j];
    if (y == x) return x;
    for (int j = 30; j >= 0 && y != x; --j) {
        if (fa[x][j] != fa[y][j]) {
            x = fa[x][j];
            y = fa[y][j];
        }
    }
    return fa[x][0];
}
int a[MAXN], k;
vector<int> vv;
int main() {
    while (~scanf("%d%d", &n, &q)) {
        init();
        int x, y;
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        for (int i = 1; i < n; i++) {
            scanf("%d%d", &x, &y);
            v[x].push_back(y);
            v[y].push_back(x);
        }
        dfs(1, 0);
        while (q--) {
            scanf("%d%d%d", &k, &x, &y);
            if (k == 0) a[x] = y;
            else {
                int pa = lca(x, y);
                vv.clear();
                vv.push_back(a[pa]);
                while (x != pa) {
                    vv.push_back(a[x]);
                    x = fa[x][0];
                }
                while (y != pa) {
                    vv.push_back(a[y]);
                    y = fa[y][0];
                }
                if (vv.size() < k) puts("invalid request!");
                else {
                    sort(vv.begin(), vv.end());
                    printf("%d\n", vv[vv.size() - k]);
                }
            }
        }
    }
	return 0;
}