題解 Children Trips
阿新 • • 發佈:2021-08-12
Description
給出一個大小為 \(n\) 的邊權全為 \(1,2\) 的帶權樹,有 \(q\) 此查詢,每次給出 \(u,v,p\) ,問 \(u\to v\) 每次可以最多走邊權和 \(\le p\) 的路徑,問最少走多少次。
\(n,q\le 10^5\)
Solution
因為自己沒有想出來,所以還是寫一發題解。
首先邊權比較有迷惑性,但是這個 \(\le 2\) 確實多少用。考慮根號分治,可以發現 \(p>\sqrt n\) 最多隻會爬 \(\sqrt n\) 次,所以可以直接暴力爬,即倍增找到每次可以走到的下一個點。
考慮 \(\sqrt n\),發現我們仍可以使用倍增,即處理出一個點走 \(2^i\)
複雜度 \(\Theta(n\sqrt n\log n)\) 。
Code
#include <bits/stdc++.h> using namespace std; #define Int register int #define MAXN 100005 template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;} template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);} template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');} template <typename T> inline void chkmax (T &a,T b){a = max (a,b);} template <typename T> inline void chkmin (T &a,T b){a = min (a,b);} int n,m,siz; struct edge{int v,w;}; vector <edge> g[MAXN]; struct str{ int u,v,P,ind; bool operator < (const str &t)const{return P < t.P;} }seq[MAXN]; int dep[MAXN],dis[MAXN],par[MAXN][21]; int getlca (int u,int v){ if (dep[u] < dep[v]) swap (u,v); for (Int i = 20,dis = dep[u] - dep[v];~i;-- i) if (dis >> i & 1) u = par[u][i]; if (u == v) return u; else{ for (Int i = 20;~i;-- i) if (par[u][i] ^ par[v][i]) u = par[u][i],v = par[v][i]; return par[u][0]; } } void dfs (int u,int fa){ dep[u] = dep[fa] + 1,par[u][0] = fa; for (Int i = 1;i <= 20;++ i) par[u][i] = par[par[u][i - 1]][i - 1]; for (edge to : g[u]){ int v = to.v,w = to.w; if (v == fa) continue; dis[v] = dis[u] + w,dfs (v,u); } } int getjump (int u,int p){ for (Int i = 20;~i;-- i) if (par[u][i] && dis[u] - dis[par[u][i]] <= p) // cout << u << " -> " << par[u][i] << endl, p -= dis[u] - dis[par[u][i]],u = par[u][i]; return u; } struct node{int lft,stp;}; node getit1 (int u,int lca,int p){ int stp = 0; while (u ^ lca){ int now = getjump (u,p); if (dep[now] > dep[lca]) stp ++,u = now; else break; } return node{dis[u] - dis[lca],stp}; } int st[MAXN][21]; void makeit (int P){ for (Int i = 1;i <= n;++ i) st[i][0] = getjump (i,P); // cout << P << ": ------------ " << endl; for (Int j = 1;(1 << j) <= n;++ j) for (Int i = 1;i <= n;++ i) st[i][j] = st[st[i][j - 1]][j - 1]; // cout << i << " jump " << (1 << j) << ": " << st[i][j] << endl; } node getit2 (int u,int lca,int p){ int stp = 0; for (Int i = 20;~i;-- i) if (st[u][i] && dep[st[u][i]] > dep[lca]) stp += (1 << i),u = st[u][i]; return node{dis[u] - dis[lca],stp}; } int solveit (int x,int y,int P){ if (x == y) return 0; int lca = getlca (x,y); node t1 = P <= siz ? getit2 (x,lca,P) : getit1 (x,lca,P); node t2 = P <= siz ? getit2 (y,lca,P) : getit1 (y,lca,P); int now = t1.stp + t2.stp; if (t1.lft + t2.lft <= P) return now + 1; else return now + 2; } int ans[MAXN]; signed main(){ // freopen ("data.in","r",stdin); // freopen ("f1.out","w",stdout); read (n),siz = sqrt (n); for (Int i = 2,u,v,w;i <= n;++ i) read (u,v,w),g[u].push_back (edge{v,w}),g[v].push_back (edge{u,w}); dfs (1,0),read (m); for (Int i = 1;i <= m;++ i) read (seq[i].u,seq[i].v,seq[i].P),seq[i].ind = i; sort (seq + 1,seq + m + 1); for (Int i = 1;i <= m;++ i){ if (seq[i].P <= siz){ if (seq[i].P != seq[i - 1].P) makeit (seq[i].P); ans[seq[i].ind] = solveit (seq[i].u,seq[i].v,seq[i].P); } else ans[seq[i].ind] = solveit (seq[i].u,seq[i].v,seq[i].P); } for (Int i = 1;i <= m;++ i) write (ans[i]),putchar ('\n'); return 0; }