题面

Sol

路径第\(k\)大

无解直接判断就好了

然后整体二分，加上树链剖分+树状数组统计

# include 
    
     # define RG register# define IL inline# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;const int _(2e5 + 5);IL ll Input(){    RG ll x = 0, z = 1; RG char c = getchar();    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);    return x * z;}int size[_], fa[_], deep[_], top[_], son[_], dfn[_], Index;int fst[_], nxt[_], to[_], cnt, n, c[_];IL void Modify(RG int x, RG int v){    for(; x <= n; x += x & -x) c[x] += v;}IL int Query(RG int x){    RG int ret = 0;    for(; x; x -= x & -x) ret += c[x];    return ret;}IL void Add(RG int u, RG int v){    nxt[cnt] = fst[u]; to[cnt] = v; fst[u] = cnt++;}IL void Dfs1(RG int u){    size[u] = 1;    for(RG int e = fst[u]; e != -1; e = nxt[e]){        if(size[to[e]]) continue;        fa[to[e]] = u; deep[to[e]] = deep[u] + 1;        Dfs1(to[e]);        size[u] += size[to[e]];        if(size[to[e]] > size[son[u]]) son[u] = to[e];    }}IL void Dfs2(RG int u, RG int Top){    dfn[u] = ++Index; top[u] = Top;    if(son[u]) Dfs2(son[u], Top);    for(RG int e = fst[u]; e != -1; e = nxt[e])        if(!dfn[to[e]]) Dfs2(to[e], to[e]);}IL int LCA(RG int u, RG int v){    while(top[u] ^ top[v]){        if(deep[top[u]] > deep[top[v]]) swap(u, v);        v = fa[top[v]];    }    return deep[u] > deep[v] ? v : u;}IL int Calc(RG int u, RG int v){    RG int ret = 0;    while(top[u] ^ top[v]){        if(deep[top[u]] > deep[top[v]]) swap(u, v);        ret += Query(dfn[v]) - Query(dfn[top[v]] - 1);        v = fa[top[v]];    }    if(deep[u] > deep[v]) swap(u, v);    ret += Query(dfn[v]) - Query(dfn[u] - 1);    return ret;}int m, tot, ans[_], Q, val[_];struct Data{    int id, x, y, k, op;} q[_], q1[_], q2[_];IL void Solve(RG int l, RG int r, RG int L, RG int R){    if(L > R) return;    if(l == r){        for(RG int i = L; i <= R; ++i)            if(q[i].id && ans[q[i].id] != -1) ans[q[i].id] = l;        return;    }    RG int mid = (l + r) >> 1, t1 = 0, t2 = 0;    for(RG int i = L; i <= R; ++i)        if(q[i].id){            RG int s = Calc(q[i].x, q[i].y);            if(s >= q[i].k) q2[++t2] = q[i];            else q[i].k -= s, q1[++t1] = q[i];        }        else{            if(q[i].y > mid) q2[++t2] = q[i], Modify(dfn[q[i].x], q[i].op);            else q1[++t1] = q[i];        }    for(RG int i = L; i <= R; ++i)        if(!q[i].id && q[i].y > mid) Modify(dfn[q[i].x], -q[i].op);    for(RG int i = L, j = 1; j <= t1; ++i, ++j) q[i] = q1[j];    for(RG int i = L + t1, j = 1; j <= t2; ++i, ++j) q[i] = q2[j];    Solve(l, mid, L, L + t1 - 1); Solve(mid + 1, r, L + t1, R);}int main(RG int argc, RG char* argv[]){    tot = n = Input(); m = Input(); RG int mx = 0;    for(RG int i = 1; i <= n; ++i){        q[i].x = i, q[i].y = val[i] = Input(), q[i].op = 1;        fst[i] = -1; mx = max(mx, val[i]);    }    for(RG int i = 1, x, y; i < n; ++i)        x = Input(), y = Input(), Add(x, y), Add(y, x);    Dfs1(1); Dfs2(1, 1);    for(RG int i = 1; i <= m; ++i){        RG int tk = Input(), tx = Input(), ty = Input();        if(tk){            RG int lca = LCA(tx, ty); ++Q;            RG int len = deep[tx] + deep[ty] - 2 * deep[lca] + 1;            if(tk > len) ans[Q] = -1;            q[++tot] = (Data){Q, tx, ty, tk, 0};        }        else{            q[++tot] = (Data){0, tx, val[tx], 0, -1};            q[++tot] = (Data){0, tx, val[tx] = ty, 0, 1};            mx = max(mx, ty);        }    }    Solve(0, mx, 1, tot);    for(RG int i = 1; i <= Q; ++i)        if(ans[i] == -1) puts("invalid request!");        else printf("%d\n", ans[i]);    return 0;}