2015 Beijing online race Couple Trees
Test instructions: Two trees, finding the nearest public ancestor of two nodes on different trees
Train of thought: The team that I saw in the game is not many, did not think carefully. Today, the problem only found that there is a multiplication algorithm, I actually do not know.
The solution comes from Qscqesze, this actually before if understand multiplication words or is not very difficult, but this question data also is not very to force, limit data theoretically can not pass.
Other solutions are tree-chain split? is not very clear. So the water is over ...
1#include <iostream>2#include <cstdio>3#include <fstream>4#include <algorithm>5#include <cmath>6#include <deque>7#include <vector>8#include <queue>9#include <string>Ten#include <cstring> One#include <map> A#include <stack> -#include <Set> - #defineLL Long Long the #defineEPS 1e-8 - #defineINF 0x3f3f3f3f - #defineMAXN 100005 - using namespacestd; + - intf1[maxn][ -], f2[maxn][ -]; + intDEEP1[MAXN], DEEP2[MAXN]; A intStep1, Step2, ans; at voidWorkintXinty) { -Step1 = Step2 =1; - while(X! =y) { - if(X <y) { - //x < y means y is not X ' s ancestor - for(inti = the; I >=0; i--){ in if(F2[y][i] >x) { -y =F2[y][i]; toStep2 + =1<<i; + Break; - } the } *y = f2[y][0]; $step2++;Panax Notoginseng } - Else{ the for(inti = the; I >=0; i--){ + if(F1[x][i] >y) { Ax =F1[x][i]; theStep1 + =1<<i; + Break; - } $ } $x = f1[x][0]; -step1++; - } the } -Ans =x;Wuyi } the intMain () - { Wu #ifndef Online_judge -Freopen ("In.txt","R", stdin); About //freopen ("OUT.txt", "w", stdout); $ #endif //Open_file - intN, M; - while(~SCANF ("%d%d", &n, &m)) { - intx, y; Adeep1[1] = deep2[1] =1; + for(inti =0; I <= the; i++){ thef1[1][i] = f2[1][i] =1; - } $ for(inti =2; I <= N; i++){ thescanf"%d", &x); thef1[i][0] =x; theDeep1[i] = Deep1[x] +1; the for(intj =1; J <= the; J + +){ -F1[I][J] = f1[f1[i][j-1]][j-1]; in } the } the for(inti =2; I <= N; i++){ Aboutscanf"%d", &x); thef2[i][0] =x; theDeep2[i] = Deep2[x] +1; the for(intj =1; J <= the; J + +){ +F2[I][J] = f2[f2[i][j-1]][j-1]; - } the }BayiAns =0; the for(inti =1; I <= m; i++){ thescanf"%d%d", &x, &y); -x = (x + ans)% n +1; -y = (y + ans)% n +1; the Work (x, y); theprintf"%d%d%d\n", ans, step1, step2); the } the } -}
2015 Beijing Network race Couple Trees multiplication algorithm