Let’s define the Eulerian traversal of a tree (a connected undirected graph without cycles) as follows: consider a depth-first search algorithm which traverses vertices of the tree and enumerates them in the order of visiting (only the first visit of each vertex counts). This function starts from the vertex number 1 and then recursively runs from all vertices which are connected with an edge with the current vertex and are not yet visited in increasing numbers order. Formally, you can describe this function using the following pseudocode:
next_id = 1
id = array of length n filled with -1
visited = array of length n filled with false
function dfs(v):
visited[v] = true
id[v] = next_id
next_id += 1
for to in neighbors of v in increasing order:
if not visited[to]:
dfs(to)
You are given a weighted tree, the vertices of which were enumerated with integers from 1 to n using the algorithm described above.
A leaf is a vertex of the tree which is connected with only one other vertex. In the tree given to you, the vertex 1 is not a leaf. The distance between two vertices in the tree is the sum of weights of the edges on the simple path between them.
You have to answer q queries of the following type: given integers v, l and r, find the shortest distance from vertex v to one of the leaves with indices from l to r inclusive.
Input
The first line contains two integers n and q (3≤n≤500000,1≤q≤500000) — the number of vertices in the tree and the number of queries, respectively.
The (i–1)-th of the following n–1 lines contains two integers pi and wi (1≤pi<i,1≤wi≤109), denoting an edge between vertices pi and i with the weight wi.
It’s guaranteed that the given edges form a tree and the vertices are enumerated in the Eulerian traversal order and that the vertex with index 1 is not a leaf.
The next q lines describe the queries. Each of them contains three integers vi, li, ri (1≤vi≤n,1≤li≤ri≤n), describing the parameters of the query. It is guaranteed that there is at least one leaf with index x such that li≤x≤ri.
Output
Output q integers — the answers for the queries in the order they are given in the input.
Examples
input
5 3 1 10 1 1 3 2 3 3 1 1 5 5 4 5 4 1 2
output
3 0 13
input
5 3 1 1000000000 2 1000000000 1 1000000000 1 1000000000 3 4 5 2 1 5 2 4 5
output
3000000000 1000000000 2000000000
input
11 8 1 7 2 1 1 20 1 2 5 6 6 2 6 3 5 1 9 10 9 11 5 1 11 1 1 4 9 4 8 6 1 4 9 7 11 9 10 11 8 1 11 11 4 5
output
8 8 9 16 9 10 0 34
Note
In the first example, the tree looks like this:
In the first query, the nearest leaf for the vertex 1 is vertex 4 with distance 3. In the second query, the nearest leaf for vertex 5 is vertex 5 with distance 0. In the third query the nearest leaf for vertex 4 is vertex 4; however, it is not inside interval [1,2] of the query. The only leaf in interval [1,2] is vertex 2 with distance 13 from vertex 4.
Solution:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 | #include <bits/stdc++.h> using namespace std; string to_string(string s) { return '"' + s + '"' ; } string to_string( const char * s) { return to_string((string) s); } string to_string( bool b) { return (b ? "true" : "false" ); } template < typename A, typename B> string to_string(pair<A, B> p) { return "(" + to_string(p.first) + ", " + to_string(p.second) + ")" ; } template < typename A> string to_string(A v) { bool first = true ; string res = "{" ; for ( const auto &x : v) { if (!first) { res += ", " ; } first = false ; res += to_string(x); } res += "}" ; return res; } void debug_out() { cerr << endl; } template < typename Head, typename ... Tail> void debug_out(Head H, Tail... T) { cerr << " " << to_string(H); debug_out(T...); } #ifdef LOCAL #define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__) #else #define debug(...) 42 #endif class segtree { public : struct node { // don't forget to set default value (used for leaves) // not necessarily neutral element! long long mn = 0; long long add = 0; void apply( int l, int r, long long v) { mn += v; add += v; } }; node unite( const node &a, const node &b) const { node res; res.mn = min(a.mn, b.mn); return res; } inline void push( int x, int l, int r) { int y = (l + r) >> 1; int z = x + ((y - l + 1) << 1); // push from x into (x + 1) and z if (tree[x].add != 0) { tree[x + 1].apply(l, y, tree[x].add); tree[z].apply(y + 1, r, tree[x].add); tree[x].add = 0; } } inline void pull( int x, int z) { tree[x] = unite(tree[x + 1], tree[z]); } int n; vector<node> tree; void build( int x, int l, int r) { if (l == r) { return ; } int y = (l + r) >> 1; int z = x + ((y - l + 1) << 1); build(x + 1, l, y); build(z, y + 1, r); pull(x, z); } template < typename M> void build( int x, int l, int r, const vector<M> &v) { if (l == r) { tree[x].apply(l, r, v[l]); return ; } int y = (l + r) >> 1; int z = x + ((y - l + 1) << 1); build(x + 1, l, y, v); build(z, y + 1, r, v); pull(x, z); } node get( int x, int l, int r, int ll, int rr) { if (ll <= l && r <= rr) { return tree[x]; } int y = (l + r) >> 1; int z = x + ((y - l + 1) << 1); push(x, l, r); node res{}; if (rr <= y) { res = get(x + 1, l, y, ll, rr); } else { if (ll > y) { res = get(z, y + 1, r, ll, rr); } else { res = unite(get(x + 1, l, y, ll, rr), get(z, y + 1, r, ll, rr)); } } pull(x, z); return res; } template < typename ... M> void modify( int x, int l, int r, int ll, int rr, const M&... v) { if (ll <= l && r <= rr) { tree[x].apply(l, r, v...); return ; } int y = (l + r) >> 1; int z = x + ((y - l + 1) << 1); push(x, l, r); if (ll <= y) { modify(x + 1, l, y, ll, rr, v...); } if (rr > y) { modify(z, y + 1, r, ll, rr, v...); } pull(x, z); } int find_first_knowingly( int x, int l, int r, const function< bool ( const node&)> &f) { if (l == r) { return l; } push(x, l, r); int y = (l + r) >> 1; int z = x + ((y - l + 1) << 1); int res; if (f(tree[x + 1])) { res = find_first_knowingly(x + 1, l, y, f); } else { res = find_first_knowingly(z, y + 1, r, f); } pull(x, z); return res; } int find_first( int x, int l, int r, int ll, int rr, const function< bool ( const node&)> &f) { if (ll <= l && r <= rr) { if (!f(tree[x])) { return -1; } return find_first_knowingly(x, l, r, f); } push(x, l, r); int y = (l + r) >> 1; int z = x + ((y - l + 1) << 1); int res = -1; if (ll <= y) { res = find_first(x + 1, l, y, ll, rr, f); } if (rr > y && res == -1) { res = find_first(z, y + 1, r, ll, rr, f); } pull(x, z); return res; } int find_last_knowingly( int x, int l, int r, const function< bool ( const node&)> &f) { if (l == r) { return l; } push(x, l, r); int y = (l + r) >> 1; int z = x + ((y - l + 1) << 1); int res; if (f(tree[z])) { res = find_last_knowingly(z, y + 1, r, f); } else { res = find_last_knowingly(x + 1, l, y, f); } pull(x, z); return res; } int find_last( int x, int l, int r, int ll, int rr, const function< bool ( const node&)> &f) { if (ll <= l && r <= rr) { if (!f(tree[x])) { return -1; } return find_last_knowingly(x, l, r, f); } push(x, l, r); int y = (l + r) >> 1; int z = x + ((y - l + 1) << 1); int res = -1; if (rr > y) { res = find_last(z, y + 1, r, ll, rr, f); } if (ll <= y && res == -1) { res = find_last(x + 1, l, y, ll, rr, f); } pull(x, z); return res; } segtree( int _n) : n(_n) { assert (n > 0); tree.resize(2 * n - 1); build(0, 0, n - 1); } template < typename M> segtree( const vector<M> &v) { n = v.size(); assert (n > 0); tree.resize(2 * n - 1); build(0, 0, n - 1, v); } node get( int ll, int rr) { assert (0 <= ll && ll <= rr && rr <= n - 1); return get(0, 0, n - 1, ll, rr); } node get( int p) { assert (0 <= p && p <= n - 1); return get(0, 0, n - 1, p, p); } template < typename ... M> void modify( int ll, int rr, const M&... v) { assert (0 <= ll && ll <= rr && rr <= n - 1); modify(0, 0, n - 1, ll, rr, v...); } // find_first and find_last call all FALSE elements // to the left (right) of the sought position exactly once int find_first( int ll, int rr, const function< bool ( const node&)> &f) { assert (0 <= ll && ll <= rr && rr <= n - 1); return find_first(0, 0, n - 1, ll, rr, f); } int find_last( int ll, int rr, const function< bool ( const node&)> &f) { assert (0 <= ll && ll <= rr && rr <= n - 1); return find_last(0, 0, n - 1, ll, rr, f); } }; int main() { ios::sync_with_stdio( false ); cin.tie(0); int n, tt; cin >> n >> tt; vector< int > p(n); vector< int > w(n); for ( int i = 1; i < n; i++) { cin >> p[i] >> w[i]; p[i]--; } vector<vector< int >> qs(n); vector< int > L(tt), R(tt), v(tt); vector< long long > res(tt, -1); for ( int i = 0; i < tt; i++) { cin >> v[i] >> L[i] >> R[i]; v[i]--; L[i]--; R[i]--; qs[v[i]].push_back(i); } vector< long long > dst(n); for ( int i = 1; i < n; i++) { dst[i] = dst[p[i]] + w[i]; } vector< int > is_leaf(n, 1); for ( int i = 1; i < n; i++) { is_leaf[p[i]] = 0; } const long long inf = ( long long ) 1e18; for ( int i = 0; i < n; i++) { if (!is_leaf[i]) { dst[i] = inf; } } segtree st(dst); vector< int > stk; vector< int > to(n, n - 1); for ( int i = 0; i < n; i++) { while (!stk.empty() && stk.back() != p[i]) { int u = stk.back(); to[u] = i - 1; stk.pop_back(); } stk.push_back(i); } debug(to); stk.clear(); for ( int i = 0; i < n; i++) { while (!stk.empty() && stk.back() != p[i]) { int u = stk.back(); st.modify(0, n - 1, -w[u]); st.modify(u, to[u], 2 * w[u]); stk.pop_back(); } stk.push_back(i); st.modify(0, n - 1, w[i]); st.modify(i, to[i], -2 * w[i]); for ( int id : qs[i]) { res[id] = st.get(L[id], R[id]).mn; } } for ( int i = 0; i < tt; i++) { cout << res[i] << '\n' ; } return 0; } |