ZOJ 3195 Design the city LCA to RMQ

Given n vertices, n-1-1 rows of u, v, and dis indicate an undirected edge and edge weight. Here a undirected tree is given.

The following m indicates m queries. Ask u v n, which is the shortest distance.

Typical conversion from LCA to RMQ

 

 

# Include <stdio. h> # include <string. h> # include <math. h ># define N 100000 # define INF 1 <29 # define Logo 17 using namespace std; inline int Min (int a, int B) {return a> B? B: a;} struct node {int f, to, dis, nex;} edge [N]; int edgenum, head [N], dis [N]; int E [N * 2], R [N], D [N * 2], en; // LCAint ST [N * 2] [Logo]; void addedge (int u, int v, int dis) {edge [edgenum]. f = u; edge [edgenum]. to = v; edge [edgenum]. dis = dis; edge [edgenum]. nex = head [u]; head [u] = edgenum ++;} void makeRmqIndex (int n, int B []) // returns the subscript {int I, j; for (I = 0; I <n; I ++) ST [I] [0] = I; for (j = 1; (1 <j) <= n; j ++) for (I = 0; I + (1 <j)-1 <n; I ++) ST [I] [j] = B [ST [I] [J-1] <B [ST [I + (1 <(J-1)] [J-1]? ST [I] [J-1]: ST [I + (1 <(J-1)] [J-1];} int LCA (int s, int v, int B []) // returns the subscript in D of the minimum value (same as the subscript in E) {s = R [s], v = R [v]; int k; if (s> v) {k = s; s = v; v = k;} k = (int) (log (v-s + 1) * 1.0) /log (2.0); return B [ST [s] [k] <B [ST [v-(1 <k) + 1] [k]? E [ST [s] [k]: E [ST [v-(1 <k) + 1] [k];} void DFS (int x, int deep) {E [en] = x; D [en] = deep; R [x] = en ++; for (int I = head [x]; I! =-1; I = edge [I]. nex) {int v = edge [I]. to; if (R [v] =-1) {dis [v] = dis [x] + edge [I]. dis; DFS (v, deep + 1); E [en] = x; D [en ++] = deep ;}} void Input (int n) {memset (head,-1, sizeof (head); edgenum = 0; while (-- n) {int u, v, dis; scanf ("% d", & u, & v, & dis); addedge (u, v, dis); addedge (v, u, dis );} memset (R,-1, sizeof (R); en = 0; dis [0] = 0;} int main () {int n, I, que, first = 0; while (~ Scanf ("% d", & n) {if (first ++) printf ("\ n"); Input (n); DFS (0, 0); makeRmqIndex (en, d); scanf ("% d", & que); while (que --) {int a, B, c; scanf ("% d ", & a, & B, & c); int ans = dis [a] + dis [B] + dis [c]-(dis [LCA (a, c, D)] + dis [LCA (B, c, D)] + dis [LCA (a, B, D)]); printf ("% d \ n", ans );}} return 0 ;}